SPHERICAL  TRIGONOMETRY 


FOR 

COLLEGES  AND  SECONDARY  SCHOOLS 


BY 


DANIEL   A.   MUjRRAY,  B.A.,  PH.D. 

INSTRUCTOR    IN    MATHEMATICS    IN    CORNELL    UNIVERSITY 
FORMERLY  SCHOLAR  AND  FELLOW  AT  JOHNS   HOPKINS  UNIVERSITY 


LONGMANS,    GRjEEN,    AND    CO. 


91  AND  93  FIFTH  AVENUE,  NEW  YORK 

LONDON  AND  BOMBAY 

1902 


COPYRIGHT,  1900, 
BY  LONGMANS,  GKEEN,  AND  CO, 


ALL  RIGHTS   RESERVED. 


PREFACE. 


THIS  book  contains  little  more  than  what  is  required  for 
the  solution  of  spherical  triangles  and  related  simple  practical 
problems.  The  articles  on  spherical  geometry  are  necessary  for 
those  who  have  not  already  studied  that  subject;  for  others, 
they  provide  a  useful  review.  More  than  usual  attention  has 
been  given  to  the  measurement  of  solid  angles.  The  explana- 
tions in  connection  with  the  astronomical  problems  are  somewhat 
fuller  than  is  customary  in  elementary  text-books  on  spherical 
trigonometry. 

I  am  indebted  to  Mr.  W.  B.  Eite,  Ph.B.,  Fellow  in  Mathe- 
matics at  Cornell  University,  for  his  kind  assistance  in  reading 
the  proof-sheets ;  and  to  Mr.  A.  T.  Bruegel,  M.M.E.,  of  the 
Pratt  Institute,  Brooklyn,  N.Y.,  for  the  pleasing  character  of 
the  diagrams. 

D.  A.  MURRAY. 

CORNELL  UNIVERSITY, 
May,  1900. 


183975 


CONTENTS. 


CHAPTER   I. 
REVIEW  OF  SOLID  AND  SPHERICAL  GEOMETRY. 

ART.  PAGE 

1-4.  Planes  and  lines  in  space.     Diedral  angles.     Solid  angles      .         .  1 

THE    SPHERE. 

5.  The  sphere  and  its  plane  sections     .......  4 

6.  Great  and  small  circles  on  a  sphere  .......  4 

7.  To  draw  circles  about  a  given  pole   .......  6 

8-9.  Proposition.    Problem 7 

10.  Lines  and  planes  which  are  tangent  to  a  sphere       ....  7 

11.  Spherical  angles 8 

ON    SPHERICAL    TRIANGLES. 

12.  Definitions 10 

13.  Propositions 12 

14.  Correspondence  between  solid  angles  and  spherical  triangles  .        .  13 

15.  Propositions      ...........  14 

16-17.   On  polar  triangles        .........  15 

18.  Definitions 18 

19.  Convention 19 

20.  Shortest  line  between  two  points  on  a  sphere 19 

PROBLEMS    OP    CONSTRUCTION. 

22.  Problems  on  great  circles      .         . '  .         .         .         .         .         .21 

23-24.   Construction  of  triangles  (six  cases)   ......  22 


CHAPTER   II. 
RIGHT-ANGLED  SPHERICAL  TRIANGLES. 

25.   Spherical  Trigonometry 28 

2(5.  Relations  between  the  sides  and  angles  of  a  right-angled  spherical 

triangle. 28 

27.  On  species 32 

vii 


viii  CONTENTS. 

ART.  PAG* 

28.  Solution  of  a  right-angled  triangle 33 

29.  The  ambiguous  case 34 

80.  Napier's  rules  of  circular  parts .35 

31.  Numerical  problems 36 

32.  Solution  of  isosceles  triangles  and  quadrantal  triangles    .         .         .39 

33.  Solution  of  oblique  spherical  triangles  (six  cases)     ....  39 

34.  Graphical  solution  of  (oblique  and  right)  spherical  triangles  .        .  42 

CHAPTER  III. 
RELATIONS  BETWEEN  THE  SIDES  AND  ANGLES  OF  SPHERICAL  TRIANGLES. 

36.  Derivation  of  the  Law  of  Sines  and  the  Law  of  Cosines  ...  44 

37.  Formulas  for  the  half -angles  and  the  half-sides        ....  47 

38.  Napier's  Analogies 50 

39.  Delambre's  Analogies  or  Gauss's  Formulas 52 

40.  Other  relations  between  the  parts  of  a  spherical  triangle  ...  53 

CHAPTER  IV. 
SOLUTION  OF  TRIANGLES. 

41.  Cases  for  solution     .        .        c        .......  54 

42.  Case  I.    Given  the  three  sides ....  ...  55 

43.  Case  II.    Given  the  three  angles 56 

44.  Case  III.     Given  two  sides  and  their  included  angle         ...  56 

45.  Case  IV.     Given  one  side  and  its  two  adjacent  angles      ...  57 

46.  Case  V.     Given  two  sides  and  the  angle  opposite  one  of  them          .  57 

47.  Case  VI.    Given  two  angles  and  the  side  opposite  one  of  them        .  61 

48.  Subsidiary  angles      .        . 61 

CHAPTER  V. 
CIRCLES  CONNECTED  WITH  SPHERICAL  TRIANGLES. 

49.  The  circumscribing  circle  .........  62 

50.  The  inscribed  circle  ....  ....  63 

51.  Escribed  circles         ..........  64 

CHAPTER  VI. 
AREAS  AND  VOLUMES  CONNECTED  WITH  SPHERES. 

52.  Preliminary  propositions 66 

53.  To  find  the  area  of  a  sphere.    Area  of  a  zone 67 


CONTENTS.  ix 

ART.  PAGE 

54.  Lunes 69 

55.  A  spherical  degree  defined        .         .         .         .        .         .         .         .09 

56.  Spherical  excess  of  a  triangle    ........  70 

57.  The  area  of  a  spherical  triangle         .         .        .         .         .         .         .71 

58.  Formulas  for  the  spherical  excess  of  a  triangle         ....  73 

59.  The  number  of  spherical  degrees  in  any  figure  on  a  sphere.     The 

spherical  excess  of  a  spherical  polygon  74 

60.  Given  the  area  of  a  figure  :  to  find  its  spherical  excess     ...  74 

61.  The  measure  of  a  solid  angle 76 

62.  The  volume  of  a  sphere 78 

63.  Definitions.     Spherical  pyramid,  segment,  and  sector      ...  78 

64.  Volume  of  a  spherical  pyramid ;  of  a  spherical  sector      ...  79 

65.  Volume  of  a  spherical  segment         .......  80 

CHAPTER  VII. 
PRACTICAL  APPLICATIONS. 

66.  Geographical  problem       .        .         .        .        .        ,        .        .        .  82 

APPLICATIONS   TO   ASTRONOMY. 

68.  The  celestial  sphere 84 

69.  Points  and  lines  of  reference  on  the  celestial  sphere         ...  86 

70.  The  horizon  system  :  Positions  described  by  altitude  and  azimuth  .  87 

71.  The  equator  system :  Positions  described  by  declination  and  hour 

angle 88 

72.  The  altitude  of  the  pole  is  equal  to  the  latitude  of  the  place  of 

observation    ...........  89 

73.  To  determine  the  time  of  day   ........  90 

74.  To  find  the  time  of  sunrise        ........  91 

75.  Theorem 92 

76.  The  equator  system :  Positions  described  by  declination  and  right 

ascension 92 

77.  The  ecliptic  system  :  Positions  described  by  latitude  and  longitude  93 

APPENDIX. 

NOTE  A.  On  the  fundamental  formulas  of  spherical  trigonometry         .  95 

NOTE  B.   Derivation  of  formulas  for  Spherical  Excess   ....  98 

QUESTIONS  AND  EXERCISES  FOR  PRACTICE  AND  REVIEW       .         .         e  101 

ANSWERS  TO  THE  EXAMPLES  .........  113 


SPHERICAL   TRIGONOMETRY. 


CHAPTER  I. 

REVIEW  OF   SOLID  AND   SPHERICAL  GEOMETRY. 

On  beginning  the  study  of  spherical  trigonometry  it  is  advisable 
to  recall  to  mind  or  learn  some  of  the  definitions  and  propositions 
of  solid  geometry.  A  clear  and  vivid  conception  of  the  principal 
properties  of  the  sphere  is  especially  necessary.  The  definitions 
and  theorems  which  will  be  used  frequently  in  the  following  pages, 
are  quoted  in  this  chapter.* 

Planes  and  Lines  in  Space.    Diedral  Angles.    Solid  Angles. 

1.  a.  Two  planes  which  are  not  parallel  intersect  in  a  straight 
line.  (Euc.  XI.  3.) 

6.  The  angle  which  one  of  two  planes  makes  with  the  other 
is  called  a  diedral  angle.  Thus,  in  Fig.  1,  the  two  planes  BD  and 


Fia.  1 


*  As  far  as  possible,  references  are  made  to  the  text  of  Euclid  ;  since,  of 
the  numerous  geometrical  text-books  in  English-speaking  countries,  his  work 
is  the  one  which  is  most  largely  used.  Those  who  use  a  text-book  other  than 
Euclid's  can  substitute  the  appropriate  references. 

1 


s/'//r./.'/r.l/-,    TRIGONOMETRY.  [Cn.  I. 


.  I  /•;  intersect  in  the  straight  line  AB,  and  form  the  diedral  angle 

/•'.  i  no. 

c.  The  planes  AE  ancM|la7  are  called  the  faces,  and  the  line 
.  I  />'  is  called  the  edge,  of  We  diedral  angle.  The  faces  are  unlim- 
ited in  extent.  The  magnitude  of  the  diedral  angle  depends,  not 
upon  tho  extent  of  its  face^rbut  only  upon  their  relative  position. 
(Just  as  the  magnitude  or  a  plane  angle  depends,  not  upon  the 
lengths  of  its  boundary  lines,  but  upon  their  relative  position.) 

cZ.  If  PR  be  drawn  perpendicular  to  AB  in  the  plane  AE,  and 
PS  .be  drawn  perpendicular  to  AB  in  the  plane  AC,  the  angle 
RPS  is  called  the  plane  angle  of  the  diedral  angle. 

e.  If  a  plane  is  drawn  perpendicular  to  the  edge  of  a  diedral 
angle,  the  intersections  of  this  plane  with  the  faces  of  the  diedral 
angle  form  the  plane  angle  of  the  diedral  angle.     (See  Euc.  XL  4.) 
Thus,  if  the  plane  M  be  passed  through  p  perpendicular  to  AB, 
the  intersections,  pr,  ps,  of  the  plane  M  and  the  planes  AE,  AC, 
form  the  angle  rps  which  is  the  plane  angle  of  FABC. 

f.  All  plane  angles  of  the  same  diedral  angle  are  equal.     (See 
Euc.  XI.  10.)     Hence,  the  plane  angle  can  be  taken  as  the  measure 
of  the  diedral  angle. 

2.  a.   If  a  straight  line  be  at  right  angles  to  a  plane,  every 
plane  which  passes  through  the  line  is  at  right  angles  to  that 
plane.     (Euc.  XI.  18.) 

b.  If  two  planes  which  cut  one  another  be  each  of  them  per- 
pendicular to  a  third  plane,  their  common  section  is  perpendicular 
to  the  same  plane.  (Euc.  XL  19.) 

3.  a.   When  three  or  more  planes  meet  in  a  common  point, 
they  are  said  to  form  a  solid  angle,  or  a  polyedral  angle,  at  that 
point. 

The  point  in  which  the  planes  meet  is  called  the  vertex  of 
the  solid  angle;  the  intersections  of  the  planes  are  called  its 
edges;  the  portions  of  the  planes  between  the  edges  are  called 
its  faces;  the  plane  angles  formed  by  the  edges  are  called  its 
face  angles;  and  the  diedral  angles  formed  'at  the  edges  by  the 
planes  are  called  the  diedral  angles  (or  the  edge  angles)  of  the 
solid  angle. 


3-4] 


GEOMETRY  OF  THE  SPHERE. 


Thus,  in  Fig.  2,  for  the  solid  angle  formed  at  8 :  the  vertex  is 
S-,  SB,  SO,  SD,  SE,  are  the  edges;  BSE,  ESD,  etc.,  are  the 
faces;  the  face  angles  are  the  angles  BSE,  ESD,  DSC,  CSB; 
the  diedral  (or  edge)  angles  are  BESD,  EDSC,  etc. 


Fio.  2 


FIG.  3 


b.  A  solid  angle  with  three  faces  is  called  a  triedral  angle. 
Thus,  the  solid  angle  at  0  (Fig.  3)  is  a  triedral  angle. 

(The  measurement  of  solid  angles  is  discussed  in  Art.  61.  The 
magnitude  of  the  solid  angle  in  nowise  depends  upon  the  lengths 
of  its  edges.) 

4.  a.  The  sum  of  any  two  face  angles  of  a  triedral  angle  is 
greater  than  the  third.  (See  Euc.  XI.  20.) 

b.  The  sum  of  the  face  angles  of  any  solid  angle  is  less  than 
four  right  angles  (Euc.  XI.  21).  (This  is  true,  in  general,  only 
when  the  polygon,  say  BE  DC  (Fig.  2),  formed  by  the  intersec- 
tions of  the  faces  with  a  cutting  plane  M,  does  not  have  a  re- 
entrant angle ;  in  other  words,  when  the  polygon  BE  DC  is 
convex. 

Geometry  of  the  Sphere. 

For  the  benefit  of  those  who  have  not  studied  the  geometry  of 
the  sphere,  proofs  of  a  few  of  its  propositions  are  either  out; 
lined,  or  given  in  detail.  Some  propositions  can  be  proved  very 
easily ;  hence,  only  their  enunciations  are  given.  Other  proper- 
ties of  the  sphere  will  be  proved  when  they  are  required.  (See 
Arts.  53,  54,  57,  62,  65.)  The  use  of  a  globe  on  which  figures  can 
be  drawn,  will  be  of  great  assistance  to  the  student.  If  such  a 
globe  is  not  at  hand,  a  terrestrial  or  celestial  globe  can  afford 
some  service. 


4  SPHERICAL   TRIGONOMETRY.  [Cn.  I. 

5.  The  sphere  and  its  plane  sections. 

a.  Definitions.     A  spherical  surface  is  a  surface  all  points  of 
which  are  equidistant  from  a  point  called  the  centre.     A  sphere  is 
a  solid  bounded  by  a  spherical  surface.     The  surface  of  a  sphere 
can  be  generated   by  the   revolution   of   a   semicircle  about   its 
diameter.     A  radius  of  a  sphere  is  a  straight  line  joining  the 
centre  to  any  point  on  the  surface.     According  to  the  definition 
of  a  sphere,  all  the  radii  of  a  sphere  are  equal.     A  diameter  of 
a   sphere   is    a    straight   line    passing    through   the   centre  and> 
terminated  at  both  ends  by  the   surface.     A  plane  section  of  a 
sphere  is  a  figure  whose  boundary  is  the  intersection  of  a  plane 
and  the  surface  of  the  sphere. 

b.  Proposition.     TJie  boundary  of  every  plane  section  of  a  sphere 
is  a  circle. 

Let  the  sphere  whose  centre  is  at  0  be  cut  by  a  plane  in  the 
section  ABD ;  then  ABD  is  a  circle.  Through  0  draw  OC  per- 
pendicular to  the  plane  ABD.  Let  A  and  B  be  any  two  points 
in  the  boundary  of  the  section  ABD.  Draw 
OA,  OB,  CA,  and  CB.  In  the  two  triangles 
OCA  and  OCB,  the  angles  at  C  are  equal 
(both  being  right  angles),  the  side  OC  is  com- 
mon, and  the  side  OA  is  equal  to  the  side  OB, 
since  both  are  radii  of  the  sphere.  Hence  the 
triangles  are  equal  in  every  respect,  and  CA 
is  equal  to  CB.  But  A  and  B  are  any  two 
points  on  the  boundary  of  the  section ;  hence 
all  points  on  the  boundary  are  equidistant  from  C.  Therefore 
ABD  is  a  circle  whose  centre  is  at  (7,  the  foot  of  the  perpen- 
dicular let  fall  from  the  centre  0  to  the  cutting  plane  ABD. 

6.  Great  and  small  circles  on  a  sphere. 

a.  Definitions.  The  section  in  which  a  sphere  is  cut  by  a  plane 
is  called  a  Great  Circle  when  the  plane  passes  through  the  centre 
of  the  sphere ;  the  section  is  called  a  Small  Circle  when  the  cut- 
ting plane  does  not  pass  through  the  centre  of  the  sphere.  Thus, 
on  a  terrestrial  globe  the  meridians  and  equator  are  great  circles ; 
the  parallels  of  latitude  are  small  circles.  The  Axis  of  a  circle  of 


5-6.]  CIRCLES  ON  A   SPHERE.  5 

a  sphere  is  the  diameter  of  the  sphere  perpendicular  to  the  plane 
of  the  circle ;  the  extremities  of  the  axis  are  called  the  Poles  of 
the  circle  and  any  of  its  arcs.  Thus,  in  Fig.  4,  Art.  5,  N  and  S 
are  the  poles  of  the  circle  ABD  and  of  the  arcs  AB  and  BD,  It 
is  obvious  that  all  circles  made  by  the  intersections  of  parallel 
planes  with  a  sphere  have  the  same  axis  and  poles.  For  instance, 
all  parallels  of  latitude  have  the  same  axis  and  poles,  namely, 
the  polar  axis  of  the  earth  and  the  North  and  South  Poles. 

6.  Propositions  relating  to  great  circles. 

Every  great  circle  bisects  the  surface  of  the  sphere;  e.g.  the 
equator  bisects  the  surface  of  a  terrestrial  globe. 

Any  two  greatf  circles  bisect  each  other;  e.g.  the  meridians 
bisect  one  another  at  the  poles.  All  great  circles  of  a  sphere  are 
equal ;  since  their  radii  are  radii  of  the  sphere. 

A  great  circle  caTi  be  passed  through  any  two  points  on  a 
sphere ;  since  a  plane  can  be  made  to  pass  through  these  two 
points  and  the  centre  of  the  sphere,  and  this  plane  intersects  the 
surface  of  a  sphere  in  a  great  circle.  In  general,  only  one  great 
circle  can  be  drawn  through  two  points  on  a  sphere,  since  these 
points  and  the  centre  determine  a  plane ;  but,  when  the  two  given 
points  are  at  the  ends  of  a  diameter  an  infinite  number  of  great 
circles  can  be  drawn  through  them;  e.g.  the  meridians  passing 
through  the  North  and  South  Poles. 

c.  Definitions.  By  distance  between  two  points  on  a  sphere  is 
meant  the  shorter  arc  of  the  great  circle  passing  through  them. 
It  is  shown  in  Art.  20  that  this  arc  is  the  shortest  line  that  can  be 
drawn  on  the  surface  of  the  sphere  from  the  one  point  to  the 
other.  For  example,  the  arc  NA  in  Fig.  4  measures  the  distance 
between  the  points  N  and  A.  [Ex.  Distance  between  N  and  S  ?] 

NOTE.  The  theorem  in  Art.  20  can  be  shown  mechanically  by  taking  two 
points  on  a  parallel  of  latitude  on  a  globe  and  letting  a  string  be  stretched 
taut  from  one  point  to  the  other.  The  string  will  not  lie  on  the  parallel,  but 
will  evidently  be  in  a  plane  which  passes  through  the  centre  of  the  sphere.  If 
the  two  points  be  on  a  meridian,  the  stretched  string  will  lie  on  the  meridian. 

By  angular  distance  between  two  points  on  a  sphere  is  meant 
the  angle  subtended  at  the  centre  of  the  sphere  by  the  arc  joining 
the  given  points.  Thus  in  Fig.  4  the  angle  NO  A  is  the  angular 
distance  of  A  from  N. 


6  srilKRICAL    TETGONOMETEY.  [On.  I. 

il.  Propositions  and  definitions  relating  to  small  and  great  circles. 

In  Fig.  4  till  tlin  arcs  of  givat  circles,  as  NA,  NB,  ND,  drawn 
from  points  on  the  circle  ABD  to  the  pole  N9  are  equal.  Thus 
the  arcs  of  meridians  on  a  terrestrial  globe  drawn  from  a  parallel 
of  latitude  to  the  North  Pole  are  equal.  The  chords  NA,  NB, 
\I>,  are  all  equal;  the  angles  AON,  hON,  DON,  are  likewise 
t'qiuil.  It  thus  appears  that  all  points  in  the  circumference  of  a 
circle  on  a  sphere  are  equally  distant  from  a  pole  of  the  circle, 
whether  the  distance  be  measured  by  the  arc  of  a  great  circle 
joining  one  of  the  points  and  the  pole,  or  by  the  straight  line 
joining  the  point  and  the  pole,  or  by  the  angle  which  such  an  arc 
or  chord  subtends  at  the  centre  of  the  sphere. 

Definitions.  The  last  mentioned  angle  is  called  the  angular 
radius  of  the  circle.  The  angular  radius  of  a  great  circle  is  evi- 
dently a  right  angle.  The  polar  distance  of  a  circle  on  a  sphere 
is  its  distance  from  its  pole,  the  distance  being  measured  along 
an  arc  of  a  great  circle  passing  through  the  pole.  Thus  the  north 
polar  distance  of  a  parallel  of  latitude  is  its  distance  from  the 
North  Pole  measured  along  a  meridian.  The  term  quadrant, 
when  used  in  connection  with  a  sphere,  usually  means  an  arc 
equal  in  length  to  one-fourth  of  a  great  circle.  The  polar  dis- 
tance of  each  point  on  a  great  circle  is  evidently  a  quadrant; 
e.g.  a  point  on  the  equator  is  at  a  quadrant's  distance  from  the 
North  or  South  Pole.  Points  on  a  great  circle  are  equidistant 
from  both  its  poles.  The  polar  distance  of  a  circle  may  be  called 
the  radius  of  the  circle. 

'  7.  To  draw  circles  upon  the  surface  of  a  sphere  about  a  given  point 
as  pole. 

(a)  With  a  pair  of  compasses.  Open  the  compasses  until  the  dis- 
tance between  the  points  of  the  compasses  is  equal  to  the  chord 
of  the  polar  distance  (or,  what  is  the  same  thing,  the  chord  sub- 
tended by  the  angular  radius)  of  the  required  circle.  Then,  one 
point  being  placed  and  kept  fixed  at  the  pole,  the  other  can  describe 
the  circle. 

(&)  With  a  string.  Take  a  string  equal  in  length  to  the  polar 
distance  of  the  required  circle.  If  the  string  be  kept  stretched 


OF 


7-10.] 


CIRCLES   ON  A   SPHERE. 


taut,  and  one  end  be  fixed  at  the  pole  while  the  other  end  moves 
on  the  sphere,  the  required  circle  will  be  described.. 

In  order  to  describe  a  great  circle  the  polar  distance  must  be 
taken  equal  to  a  quadrant  of  the  sphere. 

8.  Proposition.     If  a  point  on  the  surface  of  a  sphere  lies  at  a 
quadrant's  distance  from  each  of  two  points,  it  is  the  pole  of  the  great 
circle  passing  through  these  points. 

If  the  point  P  be  at  a  quadrant's  distance 
from  each  of  the  points  A  and  B,  then  P  is  the 
pole  of  the  great  circle  passing  through  A  and 
B.  Let  0  be  the  centre  of  the  sphere,  and  draw 
OA,  OB,  OP.  Since  PA  and  PB  are  quad- 
rants, the  angles  POA  and  POB  are  right 
angles.  Hence  PO  is  perpendicular  to  the 
plane  AOB  (Euc.  XI.  4)  ;  therefore  P  is  the 
pole  of  the  great  circle  ABL. 

9.  Problem.      Through  tivo  given  points  to  draw  an  arc  of  a  great 
circle.     About  each  point  as  a  pole  draw  a  great  circle  (Art.  7). 
The  two  points  of  intersection  of  the  great  circles  thus  drawn  are 
each  at  a  quadrant's  distance  from  the  two  given   points;  and 
hence,  by  Art.  8,  are  the  poles  of  the  great  circle  through  the  two 
given  points.     Accordingly,  the  required  arc  will  be  obtained  by 
describing  a  great  circle  about  either  of  these  poles. 

NOTE.  If  the  two  given  points  are  diametrically  opposite,  an  infinite  num- 
ber of  great  circles  can  be  drawn  through  them.  (Art.  6.  &.) 

10.  Lines  and  planes  which  are  tangent  to  a  sphere. 

a.  Definitions.  A  straight  line  or  a  plane  is  said  to  be  tangent 
to  a  sphere  when  it  has  but  one  point  in  common  with  the  surface 
of  the  sphere.  The  common  point  is  called  the  point  of  contact  or 
point  oftangency. 


.  6 


8 


SPHERICAL    TRIGONOMETRY. 


[Cn.  I. 


b.  Propositions.     (See  Fig.  6.) 

A  plane  or  a  line  perpendicular  to  a  radius  at  its  extremity  is 
tangent  to  the  sphere.  [Suggestion  for  proof :  The  perpendicular 
is  the  shortest  line  that  can  be  drawn  from  a  point  to  a  plane.] 

A  tangent  to  an  arc  of  a  great  circle  at  any  point  of  the  arc  is 
perpendicular  to  the  radius  (of  the  sphere)  drawn  to  the  point. 

11.   On  spherical  angles. 

a.  Definitions.  The  angle  made  by  any  two  curves  meeting  in 
a  common  point  is  the  angle  formed  by  the  two  tangents  to  the 

curves  at  that  point.  Thus  in  Fig.  7, 
the  angle  made  by  the  curves  Ci  and  C2 
at  the  point  P,  is  the  angle  T^TT*  be- 
tweeii  the  tangents  to  Ci  and  O2  at  P. 
(This  definition  applies  to  all  curves, 
whether  they  are  in  the  same  plane  or 
riot.) 

A  spherical  angle  is  the  angle  formed 
by  two  intersecting  arcs  of  great  circles 

on  the  surface  of  a  sphere.  Thus  the  angle  formed  by  the 
arcs  CA  and  CB  (Fig.  8)  is  a  spherical  angle.  This  angle  is 
the  angle  ECD  between  the  tangents  CE  and  CD.  But  ECD  is 
the  plane  angle  of  the  diedral  angle  between  the  planes  CO  A  and 
COB  which  are  the  planes  of  the  arcs  CA  and  CB.  Thus  the 
spherical  angle  is  equal  to  the  diedral  angle  of  the  planes  of  the  arcs 
forming  the  angle. 
C 


0i 


FIG.  7 


FIG.  9 


FIG. 


b.  Propositions.  (1)  If  two  arcs  of  great  circles  intersect,  the 
opposite  vertical  angles  thus  formed  are  equal.  Thus  in  Fig.  49, 
Art.  57,  the  angles  BAC  and  B'AC1  are  equal. 


11.]  SPHERICAL  ANGLES.  9 

(2)  If  one  arc  of  a  great  circle  meets  another  arc  of  a  great 
circle,  the  sum  of  the  adjacent  spherical  angles  is  equal  to  two 
right  angles.  Thus  in  Fig.  49,  CAB  +  CAB'  =  2  right  angles. 

NOTE.  It  is  shown  in  plane  geometry  that  angles  at  the  centre  of  a 
circle  are  proportional  to  their  intercepted  arcs  ;  hence,  the  angles  can  be 
measured  by  the  arcs.  Accordingly,  if  each  right  angle  at  the  centre  of  a 
circle  (Fig.  9)  be  divided  into  90  equal  parts  called  degrees,  and  the  circle  be 
divided  into  360  equal  parts,  also  called  degrees,  then  the  number  of  degrees 
(of  angle)  in  any  angle  AOB  is  equal  to  the  number  of  degrees  (of  arc)  in  AB, 
the  arc  subtended  by  AOB.  [When  it  is  necessary  to  distinguish  between 
dryrc.es  of  angle  and  degrees  of  arc,  the  former  may  be  called  angular 
degrees;  and  the  latter  arcual  degrees.'] 

c.  Proposition.    A  spherical  angle  is  measured  by  the  arc  of  a 
great  circle  described  with  its  vertex  as  a  pole  and  included  between 
its  boundary  arcs,  produced  if  necessary  : 

Let  ABC  and  AB'C  be  two  intersecting  arcs  of  great  circles 
on  the  sphere  S  whose  centre  is  at  0.  Pass  the  plane  BOB' 
through  0  perpendicular  to  AC,  and  let  this 
plane  intersect  the  planes  ABC  and  AB'C 
in  the  radii  OB  and  OB',  and  intersect  the 
sphere  in  the  great  circle  B'BL.  From  the 
construction,  A  is  the  pole  of  the  great 
circle  B'BL.  By  Art.  1.  e.  BOB'  is  the 
plane  angle  of  the  diedral  angle  BACB', 
and,  accordingly  (Art.  11.  a),  is  equal  to  the 
spherical  angle  BAB'.  Now,  by  the  pre- 
ceding note,  the  number  of  degrees  in  the 

arc  BB'  is  equal  to  the  number  of  degrees  in  the  angle  BOB'. 
Hence,  the  number  of  degrees  in  the  arc  BB'  is  equal  to  the  num- 
ber of  degrees  in  the  angle  BAB'.  In  other  words,  the  spherical 
angle  BAB'  is  measured  by  the  arc  BB'  of  which  A  is  the  pole. 

This  can  be  illustrated  on  a  terrestrial  globe.  For  instance,  the  angle  at 
the  North  Pole  between  the  meridians  of  Paris  and  New  York  is  76°  2'  25.5"  ; 
and  this  is  the  number  of  degrees  of  arc  intercepted  by  these  meridians  on 
the  equator. 

d.  The  great  circles  drawn  through  any  point  on  a  sphere  are 
perpendicular  to  the  great  circle  of  which  the  point  is  the  pole. 


10  SPHERICAL   TRIGONOMETRY.  [Cn.  I. 

For  instance,  the  meridians  of  longitude  cross  the  equator  at 
right  angles. 

e.  The  distance  of  any  point  on  the  surface  of  a  sphere,  from  a 
circle  traced  thereon,  is  measured  by  the  shorter  arc  of  a  great 
circle  passing  through  the  point  and  perpendicular  to  the  given 
circle ;  that  is,  by  the  shorter  arc  of  the  great  circle  passing 
through  the  given  point  and  the  pole  of  the  given  circle.  For 
example,  on  a  globe  the  latitude  of  any  place  (i.e.  its  distance  in 
degrees  from  the  equator)  is  measured  by  the  arc  of  the  meridian 
intercepted  between  the  place  and  the  equator. 

N.B.  When  an  arc  on  a  sphere  is  referred  to,  an  arc  of  a  great  circle  is 
meant,  unless  expressly  stated  otherwise. 

ON   SPHERICAL  TRIANGLES. 

12.  Definitions.  A  spherical  polygon  is  a  portion  of  the  surface 
of  a  sphere  bounded  by  three  or  more  arcs  of  great  circles.  The 
bounding  arcs  are  the  sides  of  the  polygon ; 
the  points  of  intersection  of  the  sides  are 
the  vertices  of  the  polygon,  and  the  angles 
which  the  sides  make  with  one  another  are 
the  angles  of  the  polygon.  A  diagonal  of 
a  spherical  polygon  is  an  arc  of  a  great 
circle  joining  any  two  vertices  which  are 
not  consecutive. 

A  spherical  triangle  is  a  spherical  poly- 
FlG  n  gon  of  three  sides. 

Thus,  in  Fig.  11,  ABCD  is  a  spherical 

polygon;  its  sides  ate  AB,  BC,  CD,  DA',  its  angles  are  ABC, 
BCD,  CD  A,  DAB;  its  diagonals  are  BD  and  AC;  ADC  and 
ABC  are  spherical  triangles.  Since  the  sides  of  a  spherical 
polygon  are  arcs  of  great  circles,  their  magnitudes  are  expressed 
in  degrees.*  The  lengths  of  the  sides  -can  be  calculated  in  terms 
of  linear  units  when  the  radius  of  the  sphere  is  known. 

A  spherical  triangle  is  right-angled,  oblique,  scalene,  isosceles,  or 
equilateral,  in  the  same  cases  as  a  plane  triangle.  The  notation 

*  The  reason  for  expressing  the  sides  of  spherical  polygons  in  degrees  is 
considered  more  fully  in  Art.  14. 


12.]  SPHERICAL    TEIANGLES.  11 

adopted  in  discussing  the  plane  triangle  will  be  used  for  the 
spherical  triangle ;  namely,  the  triangle  will  be  denoted  by  ABC, 
and  the  sides  opposite  the  angles  A,  B,  C,  will  be  denoted  by 
a,  b,  c,  respectively. 

Two  spherical  polygons  are  equal  if  they  can  be  applied  one  to 
the  other  so  as  to  coincide.  They  are  said  to  be  symmetrical  when 
the  sides  and  angles  of  the  one  are  respectively  equal  to  the  sides 
and  angles  of  the  other,  but  arranged  in  the  reverse  order. 


Thus,  the  spherical  triangles  ABC  and  ^41S1(71  (Fig.  12)  are 
equal  if  they  can  be  brought  into  coincidence,  say,  by  sliding  one 
of  them,  as  ABC,  over  the  surface  of  the  sphere  until  it  exactly 
covers  the  surface  A-^B^C^.  Accordingly,  it  is  evident  that  if 
these  triangles  are  equal,  the  angles  A,  B,  (7,  are  respectively 
equal  to  the  angles  Alt  Bl}  Clt  and  the  sides  a,  6,  c,  are  respectively 
equal  to  the  sides  alt  &1?  c^*  On  the  other  hand,  the  triangles 
ABC  and  A2B.2C2  are  symmetrical  if  the  angles  A,  B,  C,  are  re- 
spectively equal  to  the  angles  A2,  B2,  C2,  and  the  sides  a,  6,  c,  to 
the  sides  a2,  62,  c2.  In  this  case,  the  triangle  ABC  cannot  be 
brought  into  coincidence  with  A2B2C2  by  a  sliding  motion  over 
the  surface  of  the  sphere. 

NOTE  1.  Two  symmetrical  spherical  triangles  can  be  brought  into  coinci- 
dence if  the  surface  be  covered  very  thinly  with  some  flexible  material.  For 
then  ABC  can  be  lifted  up,  turned  over,  and  the  surface  bent  (or  made  to 
'spring  back')  in  the  opposik  direction  ;  after  this  treatment,  ABC  can  be 
made  to  coincide  with  A»BzCz» 

NOTE  2.  The  meaning  of  the  phrase  reverse  order  can  be  seen  clearly 
on  considering  the  triangles  AiBid  and  A2B2C2  above.  In  AiBiCi,  on 


*  Some  of  the  sets  of  minimum  conditions  necessary  for  equality  of  spheri- 
cal triangles  are  stated  in  Art.  13. 


12  SPHERICAL   TRIGONOMETRY.  [Cn.  I. 

going  from  AI  to  B\,  thence  to  (7i,  and  thence  to  A\,  one  goes  around  any 
point,  within  the  triangle  in  a  counter-clockwise  direction.  In  AzB^Cz,  on 
the  other  liaml,  on  taking  the  respective  equal  angles  in  the  same  order  as 
before,  that  is,  on  going  from  A>2  to  B»,  thence  to  C->,  and  thence  to  A2,  one 
goes  round  any  point  within  the  triangle  A^B^Cz  in  a  clockwise  direction. 
The  directions  are  indicated  by  the  arrows. 

13.  Propositions.  (1)  Two  spherical  triangles  which  are  on  the 
same  sphere,  or  on  equal  spheres,  and  whose  parts  are  in  the  same 
order  (as  ABO  and  A^B^C^  Fig.  12)  are  equal  under  the  same  con- 
ditions as  plane  triangles,  viz. : 

(a)  When  two  sides  and  their  included  angle  in  the  one  triangle 
are  respectively  equal  to  two  sides  and  their  included  angle  in 
the  other ; 

(6)  When  a  side  and  its  two  adjacent  angles  in  the  one  triangle 
are  respectively  equal  to  a  side  and  its  two  adjacent  angles  in  the 
other ; 

(c)  When  the  three  sides  of  the  one  triangle  are  respectively 
equal  to  the  three  sides  of  the  other. 

[SUGGESTION  FOR  PROOFS.  Equality  can  be  shown  by  the  same  methods 
as  in  plane  geometry.] 

(2)  Two  spherical  triangles  which  are  on  the  same  sphere,  or  on 
equal  spheres,  and  whose  parts  are  in  the  reverse  order  (as  ABC  and 
A2B.2C2,  Fig.  12),  are  symmetrical  under  the  conditions  (a),  (6),  (c)"," 
above. 

[SUGGESTIONS  FOR  PROOF.  Construct*  a  triangle  A\B\C\  which  is  sym- 
metrical to  AvBvC*.  Under  the  given  conditions,  according  to  the  preceding 
proposition,  ABC  and  A\BiCi  have  all  their  parts  respectively  equal,  and 
hence  ABC  and  A^BzCz  have  all  their  parts  respectively  equal,  and  are 
accordingly  symmetrical.  ] 

On  a  plane  two  triangles  may  have  three  angles  of  the  one 
respectively  equal  to  three  angles  of  the  other  and  yet  not  be 
equal.  On  the  other  hand,  as  will  be  made  apparent  in  Arts. 
16,24: 

(3)  On  the  same  sphere,  or  on  equal  spheres,  two  triangles  which 
have  three  angles  of  the  one  respectively  equal  to  three  angles  of  the 
other,  are  either  equal  or  symmetrical. 

*  For  the  construction  of  spherical  triangles  under  various  conditions,  see 
Art.  24. 


13-14.]      TRIEDRAL  ANGLES:   SPHERICAL   TRIANGLES.       13 

14.  Correspondence  between  the  face  angles  and  the  diedral  angles 
of  a  triedral  angle  on  the  one  hand,  and  the  sides  and  angles  of  a 
spherical  triangle  on  the  other. 

B' 


FiO.  13 

Take  any  triedral  angle  0-A'B!Cf;  let  a  sphere  of  any  radius, 
OA  say,  be  described  about  0  as  centre ;  and  let  the  intersections 
of  this  sphere  with  the  faces  OA'B',  OB'C',  and  OC'A',  be  the 
arcs  AB,  BC,  and  CA  respectively.  The  sides  of  the  spherical 
triangle  ABC,  namely,  AB,  BC,  CA,  measure  the  face  angles, 
AOB,  BOG,  COA,  of  the  solid  angle  0-A'B'C'  (Art.  11.  b,  Note). 
By  Art.  11  the  angles  CAB,  ABC,  BCA,  of  the  spherical  triangle 
ABC  are  the  diedral  angles  between  the  planes  of  the  sides,  that 
is,  the  diedral  angles  of  the  solid  angle  0-A'B'C'. 

Hence,  to  find  the  relations  existing  between  the  face  angles 
and  the  edge  angles  of  a  triedral  angle,  is  the  same  thing  as  to 
find  the  relations  between  the  sides  and  angles  of  the  spherical 
triangle,  intercepted  by  the  faces,  upon  the  surface  of  any  sphere 
whose  centre  is  at  the  vertex  of  the  triedral  angle. 

NOTE  1.  The  number  of  degrees  in  the  intercepted  arcs  does  not  depend 
upon  the  radius  of  the  sphere.  Thus,  in  Fig.  13,  if  a  sphere  is  described 
with  a  radius  OAi,  about  0  as  a  centre,  the  number  of  degrees  in  the  inter- 
cepted arc  AiBi  is  the  same  as  the  number  of  degrees  in  the  intercepted  arc 
AB,  for  each  number  is  the  same  as  the  number  of  degrees  in  the  angle 
A' OB'. 

Since  the  face  angles  and  diedral  angles  of  a  triedral  angle  are  not  altered 
by  varying  the  radius  of  the  sphere,  the  relations  between  the  sides  and 
angles  of  the  corresponding  spherical  triangle  are  independent  of  the  length 
of  the  radius. 


14  SPHERICAL   TRIGONOMETRY.  [Cn.  I. 

NOTE  2.  Since  the  side  of  a  spherical  triangle  measures  the  angle  sub- 
tended by  it  at  the  centre,  the  side  is  measured  in  degrees  or  radians.  (See 
Art.  12.)  By  "  sin  AB,"  for  example,  is  meant  the  sine  of  the  angle  AOB, 
subtended  by  AB  at  the  centre  0. 

NOTE  3.  A  three-sided  spherical  figure,  one  or  more  of  whose  sides  is  not 
an  arc  of  a  great  circle,  is  not  regarded  as  a  spherical  triangle.  For  exam- 
ple, the  figure  bounded  by  an  arc  of  a  parallel  of  latitude  and  the  arcs  of  two 
meridians  does  not  correspond  to  a  triedral  angle  at  the  centre  of  the  sphere, 
and  is  not  a  spherical  triangle  as  defined  in  Art.  12. 

NOTE  4.  A  triedral  angle,  and  its  corresponding  spherical  triangle,  can  be 
easily  constructed.  From  stiff  cardboard  cut  out  a  circular  sector  having 
any  arc  between  0°  and  360°.  On  this  sector  draw  any  two  radii,  taking 
care,  however,  that  no  one  of  the  three  sectors  thus  formed  shall  be  greater 
than  the  sum  of  the  other  two.  Along  these  radii  cut  the  cardboard  partly 
through.  Bend  the  two  outer  sectors  over  until  their  edges  meet ;  a  figure 
like  0-ABC  (Fig.  13)  will  be  obtained.  (Find  what  happens  if  the  above 
precaution  in  drawing  the  radii  is  not  taken.) 

This  perfect  correspondence  between  the  sides  and  angles  of  a 
spherical  triangle  on  the  one  hand,  and  the  face  angles  and  die- 
dral  angles  of  the  solid  angle  subtended  at  the  centre  of  the 
sphere  by  the  triangle  on  the  other  hand,  is  very  important,  both 
for  the  deduction  of  the  relations  between  these  sides  and  angles 
and  for  the  solution  of  practical  problems.  This  correspondence 
holds  in  the  case  of  any  spherical  polygon  and  the  solid  angle 
subtended  by  it  at  the  centre  of  the  sphere.  (The  student  may 
inspect  Fig.  11.)  Hence,  from  any  property  of  polyedral  angles  an 
analogous  property  of  spherical  polygons  can  be  inferred,  and  vice 
versa. 

15.  Propositions.  (1)  Any  side  of  a  spherical  triangle  is  less 
than  the  sum  of  the  other  tivo  sides.  This  follows  from  Arts.  14 
and  4.  a. 

COR.  Any  side  of  a  spherical  polygon  is  less  than  the  sum  of 
the  remaining  sides. 

(2)  The  sum  of  the  sides  of  a  spherical  polygon  (not  re-entrant) 
is  less  than  360°.  In  other  words :  Tlie  perimeter  of  any  (non-re- 
entrant) spherical  polygon  is  less  than  the  length  of  a  great  circle. 
This  important  proposition  follows  from  Arts.  14  and  4.  b. 


15-16.]  POLAR   TRIANGLES.  15 

(3)  In  an  isosceles  spherical  triangle  the  angles  opposite  the 
equal  sides  are  equal. 

(4)  .The  arc  of  a  great  circle  drawn  from  the  vertex  of  an 
isosceles  spherical  triangle  to  the  middle  of  the  base  is  perpen- 
dicular to  the  base,  and  bisects  the  vertical  angle. 

(5)  If  two  angles  of  a  spherical  triangle  are  unequal,  the  oppo- 
site sides  are  unequal,  and  the  greater  side  is  opposite  the  greater 
angle. 

COR.  If  two  edge  angles  of  a  triedral  angle  are  unequal,  the 
opposite  face  angles  are  unequal,  and  the  greater  face  angle  is 
opposite  the  greater  diedral  angle. 

(6)  If  two  sides  of  a  spherical  triangle  are  unequal,  the  oppo- 
site angles  are  unequal,  and  the  greater  angle  is  opposite  the 
greater  side. 

Ex.     Give  the  corresponding  proposition  for  a  triedral  angle. 

Propositions  (3)— (6)  can  be  proved  in  the  same  way  as  the  cor- 
responding propositions  in  plane  geometry. 

ON    POLAR    TRIANGLES. 

16.  a.  NOTE.  Three  straight  lines  on  a  plane,  no  two  of  which  are 
parallel,  intersect  in  three  points,  and  form  one  triangle.  Three  great  circles 
on  a  sphere  have  six  points  of  intersection,  and  form  eight  spherical  triangles. 
Thus,  on  a  globe,  the  equator  and  any  two  great  circles  through  the  poles 
have  as  intersections  the  two  poles  and  the  four  points  where  the  two  great 
circles  cross  the  equator  ;  and  there  are  eight  triangles  formed,  namely,  four 
in  the  northern  hemisphere  and  four  in  the  southern. 

6.  Definitions.  If  great  circles  be  described  with  the  vertices 
of  a  spherical  triangle,  say  ABC  (Fig.  14),  as  poles ;  and  if  there 
be  taken  that  intersection  of  the  circles  described  with  B  and  C 
as  poles  which  lies  on  the  same  side  of  BC  as  does  A,  namely  A^\ 
and  similarly  for  the  other  intersections  ;  then  a  spherical  triangle 
is  formed,  which  is  called  the  polar  triangle  of  the  first  tri- 
angle ABC. 

Two  spherical  polygons  are  mutually  equilateral  when  the  sides 
of  the  one  are  respectively  equal  to  the  sides  of  the  other, 
whether  taken  in  the  same  or  in  the  reverse  order ;  the  polygons 


16 


N ."///:  /,'  1C  A  L    77,'  /  G  ONOMETE  Y. 


[Cn.  I. 


are  mutually  equiangular  when  the  angles  of  the  one  are  respec- 
tivcly  equal  to  the  angles  of  the  other,  whether  taken  in  the  same 
<>r  in  the  reverse  order. 


B 


FIG.  u 

c.  Proposition.     If  the  first  of  two  spherical  triangles  is  the  polar 
triangle  of  the  second,  then  the  second  is  the  polar  triangle  of  the  first. 

If  A'B'C'  (Fig.  14)  is  the  polar  triangle  of  ABC,  then  ABC  is 
the  polar  triangle  of  A'B'C'.  Since  A  is  the  pole  of  the  arc 
B'C',  the  point  A  is  a  quadrant's  distance  from  B'.  Also,  since  C 
is  the  pole. of  B'A,  the  point  C  is  a  quadrant's  distance  from  B'. 
Since  B'  is  thus  a  quadrant's  distance  from  both  A  and  C,  it  is 
the  pole  of  the  arc  AC  (Art.  8).  Similarly  it  can  be  shown  that 
A'  is  the  pole  of  the  arc  BC,  and  that  C'  is  the  pole  of  the  arc 
AB.  Hence  ABC  is  the  polar  triangle  of  A'B'C', 

d.  Proposition.     In  two  polar  triangles,  each  angle  of  the  one  is 
the  supplement  of  the  side  opposite  to  it  in  the  other. 

Let  ABC  and  A'pO  (Fig.  15)  be  a 
pair  of  polar  triangles,  in  which  A,  B,  C, 
A,  B'}  C',  are  the  angles,  and  a,  b,  c, 
a',  I',  c',  are  the  sides.'  Then 

A  =  180  -  a',  A  =  180  -  a, 
B  =  180  -  b',  B1  =  180  -  &, 
C  =  180  -  c',  C'  =  180  -  c. 

Produce  the  arcs  AB  and  AC  to  meet 
B'C'  in  L  and  M  respectively. 


(6-17.]  SUM  OF  ANGLES   OF  A    TR]  ANGLE.  17 

Since  B'  is  the  pole  of  ACM,  B'M  =  90°  ;  and 

since  C'  is  the  pole  of  ABL,  LC'  =  90°. 

Hence  B'M  +  LC'  =  180°  ; 

that  is,  B'M  +  MC'+LM  =  1SQ°, 

or  jB'C"  +  7v^=180°.  (1) 


Since  A  is  the  pole  of  the  arc  B'C',  the  arc  LM  measures  the 
angle  A  (Art.  11.  c). 

Hence,  (1)  becomes  A  +  a'  =  180°,  or  A  =  180°  -a'. 
The  other  relations  can  be  proved  in  a  similar  manner. 

COR.  If  two  spherical  triangles  are  mutually  equiangular, 
their  polar  triangles  are  mutually  equilateral.  If  two  spherical 
triangles  are  mutually  equilateral,  their  polar  triangles  are 
mutually  equiangular. 

NOTE.  On  account  of  the  properties  in  (cZ),  a  triangle  and  its  polar  are 
sometimes  called  supplemental  triangles. 

e.  The  use  of  the  polar  triangle.  Because  of  the  fact  that  the  sides  and 
angles  of  a  triangle  are  respectively  supplementary  to  the  angles  and  sides  of 
its  polar  triangle,  many  relations  can  be  easily  derived  by  reference  to  the 
polar  triangle.  For,  if  a  relation  is  true  for  spherical  triangles  in  general, 
then  it  is  true  for  the  polar  of  any  triangle.  Let  the  relation  be  stated  for  the 
polar  triangle  ;  in  this  statement  express  the  values  of  the  sides  and  angles  of 
the  polar  triangle  in  terms  of  the  angles  and  sides  of  the  original  triangle  ;  the 
statement  thus  derived  expresses  a  new  relation  between  the  parts  of  the 
original  triangle.  This  will  be  exemplified  in  later  articles. 

17.  Proposition.  The  sum  of  the  angles  of  a  spherical  triangle  is 
greater  than  two,  and  less  than  six,  right  angles. 

'4       Let  ABC  be  any  spherical  triangle  ;  it  is  required  to  show  that 


Construct  the  polar  triangle  A'B'C'.     Then,  by  Art.  16.  d, 
A  -f  a'  =  180°,  B  +  V  =  180°,  C  +  c'  =  180°. 


18  SPHERICAL   TRIGONOMETRY.  [Cn.  I. 

Hence,  on  adding,  A  +  B  +  C+a'  +  &'  +c'  =  540°, 
or,  .-1  +  1*  +  Cy  =  540°  -  (a'  +  6'  +  c'). 

Now  [Art.  15  (2)]  a'  +  &'  +  c'  is  less  than  360°,  and.  greater 
than  0°. 

.-.  (A  +  B  +  C)=  540°  -  (something  less  than  360°  and  greater 
than  0°). 

...  A  +  B  +  C >  540° -  360°,  i.e.  A+B  +  C>  180° ; 
and  A  +  B+  C<  540°  -  0°,  i.e.  A  +  B+C<  540°. 

18.  Definitions,  a.  The  amount  by  which  the  sum  of  the 
three  angles  of  a  spherical  triangle  is  greater  than  180°  is  called 
its  spherical  excess.  It  is  shown  in  Art.  57  that  the  area  of  a 
triangle  depends  upon  its'  spherical  excess. 

6.  A  spherical  triangle  may  have  two  right  angles,  three  right 
angles,  two  obtuse  angles,  or  three  obtuse  angles.  For  example, 
on  a  globe  the  spherical  triangle  bounded  by  any  arc  (not  90°)  on 
the  equator  and  the  arcs  of  the  meridians  joining  the  extremities 
of  the  former  arc  to  the  North  Pole,  has  two  right  angles ;  if  the 
arc  on  the  equator  is  a  quadrant,  the  triangle  •  has  three  right 
angles.  The  polar  of  the  triangle  whose  sides  are  35°,  25°,  15°, 
has  three  obtuse  angles.  A  spherical  triangle  having  two  right 
angles  is  called  a  bi-rectangular  triangle,  and  a  spherical  triangle 
having  three  right  angles  is  called  a  tri-rectangular  triangle.  A 
triangle  having  one  side  equal  to  a  quadrant  is  called  a  quadrantal 
triangle;  one  having  two  sides  each  a  quadrant  is  said  to  be 
bi~quadrantal,  and  one  having  each  of  its  three  sides  equal  to  a 
quadrant  is  said  to  be  tri-quadrantal. 

c.  A  lune  is  a  spherical  surface  bounded  by  the  halves  of  two 
great  circles.  The  angle  of  the  lune  is  the  angle  made  by  the  two 
great  circles.  For  instance,  on  a  globe  the  surface  between  the 
meridians  10°  W.  and  40°  W.  is  a  lune ;  the  angle  of  this  lime  is 
equal  to  30°.  On  the  same  circle  or  on  equal  circles  lunes  having 
equal  angles  are  equal.  (For  they  can  evidently  be  made  to 
coincide.) 


18-20.] 


SHORTEST  LINE  ON  A   CIRCLE. 


19 


FIG.  16 


19.  On  the  convention  that  each  side  of  a  spherical  triangle  be  less 
than  180°»     in  spherical  geometry  and  trigonometry  it  is  found  convenient 

to  restrict  attention  to  triangles  the  sides  of  which 
are  each  less  than  a  semicircle  or  180°.  (This  con- 
vention can  be  set  aside  when  it  is  necessary  to  con- 
sider what  is  called  the  general  spherical  triangle, 
in  which  an  element  may  have  any  value  from  0° 
to  360°.)  A  triangle  such  as  ADBC  (Fig.  16)  which 
has  a  side  ADB  greater  than  180°,  need  not  be  con- 
sidered ;  for  its  parts  can  be  immediately  deduced 
from  the  parts  of  ACS,  each  of  whose  sides  is  less 
than  180°.  It  is  easily  proved  that  if  an  angle  of  a 
spherical  triangle  is  greater  than  180°,  the  opposite 
side  is  also  greater  than  180°,  and  vice  versa.  Thus, 

in  the  triangle  ADBC,  if  the  angle  ACB  is  greater  than  180°,  so  is  the  side 
ADB  ;  and  it  ADB  is  greater  than  180°,  so  is  the  opposite  angle.  [Sugges- 
tion for  proof :  Produce  the  arc  AC  to  meet  the  arc  ADB.~] 

20.  Proposition.     The  shortest  line  that  can  be  drawn  on  the  sur- 
face of  a  sphere  between  two  given  points  is  the  arc  of  a  great  circle, 
not  greater  than  a  semicircle,  which  joins  the  points. 

Let  A  and  B  be  any  two  points  on  a  sphere,  and  let  ACB  be  a 
great-circle  arc  not  greater  than  a  semicircle;  then  ACB  is  the 
shortest  line  that  can  be  drawn  from  A  to  B  on  the  sphere. 

About  A  as  a  pole  describe  a  circle  DCE 
with  radius  AC,  and  about  B  as  a  pole 
describe  a  circle  FCG  with  radius  BC.  It 
will  be  shown  (1)  that  C  is  the  only  point 
which  is  common  to  both  these  circles; 
(2)  that  the  shortest  line  that  can  be  drawn 
from  A  to  B  on  the  surface  must  pass 
through  C. 

(1)  Take  any  point  G,  other  than  C,  on 
the  circle  FCG.  Draw  the  great-circle  arcs 
AEG  and  BG.  By  Art.  15  (1), 

AG+GB>AB;  i.e.  AG  +  GB  >  AC  +  CB. 
Now  AE  =  AC,  and  GB  =  CB. 

Hence  AE  +  GB  =  AC  +  CB ; 

and,  accordingly,  AG  >  AE. 


Fio.  17 


20  SPHERICAL   TRIGONOMETRY.  [Cn.  I. 

Therefore  G  is  outside  of  the  circle  DCE.  But  G  is  any  point 
(other  than  Gy)  on  the  circle  FOG.  Hence  C  is  the  only  point 
common  to  the  circles  DCE  and  FOG. 

(2)  Let  ADFB  be  any  line  drawn  on  the  surface  from  A  to 
J3,  but  not  passing  through  C.  Whatever  the  character  of  the 
line  AD  may  be,  a  line  exactly  like  it  can  be  drawn  from  A  to  (7; 
and  a  line  like  BF  can  be  drawn  from  B  to  C. 

[This  can  be  seen  by  regarding  A-DCE  as  a  cap  fitting  closely  to  the 
sphere,  and  supposing  that  this  cap  revolves  about  A  until  D  is  at  C.  Then 
a  line  exactly  like  AD  is  drawn  from  A  to  C.] 

These  lines  being  drawn,  there  will  be  a  line  from  A  to  B 
which  is  less  than  ADFB  by  the  part  DF.  It  has  thus  been 
proved  that  a  line  can  be  drawn  from  A  to  B  through  C  which  is 
shorter  than  any  other  line  from  A  to  B  which  does  not  pass 
through  C.  But  C  is  any  point  on  the  great-circle  arc  from  A  to 
B.  Hence  the  shortest  line  from  A  to  B  must  pass  through  every 
point  in  ACB,  and,  accordingly,  must  be  the  arc  ACB  itself. 

NOTE.  This  proposition  can  also  be  proved  by  the  method  of  limits.  It  is 
shown  that  the  length  of  any  arc  on  a  sphere  is  equal  to  the  limit  of  the 
sum  of  the  lengths  of  an  infinite  number  of  infinitesimal  great-circle  arcs 
inscribed  in  the  given  arc.  (See  Rouche"  et  De  Comberousse,  Traite  de 
Geometric.)  See  Art.  6.  c. 

PROBLEMS   OF   CONSTRUCTION. 

21.  The  actual  making  of  the  following  constructions  will  add 
much  to  the  clearness  and  vividness  of  the  notions  of  most  stu- 
dents about  the  surface  of  a  sphere.  An  easy  familiarity  with 
the  problems  of  Arts.  23,  24,  which  discuss  the  construction  of 
triangles,  will  place  the  student  in  an  advantageous  position  with 
respect  to  spherical  trigonometry.  This  position  is  similar  to 
that  occupied  by  him,  through  his  knowledge  of  the  construction 
of  plane  triangles,  when  he  entered  upon  the  study  of  plane 
trigonometry.  (See  Plane  Trigonometry,  p.  20,  Note,  Art.  21, 
Art.  34  (to  Case  I.),  Art.  53.) 

N.B.  The  student  should  try  to  make  these  constructions  for  himself, 
and  should  fall  back  upon  the  book  only  as  a  last  resort. 


21-22.]  PROBLEMS   ON  GREAT  CIRCLES.  21 

22.   Problems  on  great  circles. 

(1)  To  find  the  poles  of  a  given  great  circle.     About  any  two 
points  of  the  given  circle  as  poles,  describe  great  circles ;  their 
intersections  will  be  the  poles  required  (Art.  8). 

(2)  To  draw  a  great  circle  through  two  given  points.     About  the 
two  given  points  as  poles,  describe  great  circles ;  about  either  of 
the  intersections  of  these  circles  as  a  pole,  describe  a  great  circle ; 
this  will  be  the  circle  required.     (See  Arts.  8,  9.) 

(3)  To  cut  from  a  great  circle  an  arc  n°  long.     Separate  the 
points  of  the  compasses  by  a  distance  equal  to  a  chord  which 
subtends  a  central  angle  of  n°  in  a  circle  whose  radius  is  equal  to 
the  radius  of  the  sphere ;  place  the  points  of  the  compass  on  the 
great  circle ;  the  intercepted  arc  will  be  the  one  required. 

(A)  To  draiv  a  great  circle  through  a  given  point  perpendicular  to 
a  given  great  circle.  Find  a  pole  of  the  given  circle  by  (1) ;  draw 
a  great  circle  through  this  pole  and  the  given  point  by  '(2) ;  this 
circle  will  be  the  one  required  (Art.  11.  d). 

(5)  To  construct  a  great  circle  making  a  given  angle  with  a  given 
great  circle,  the  point  of  intersection  being  given.     About  the  given 
point  of  intersection  as  pole,  describe  a  great  circle  5  on  this  circle 
lay  off  an  arc,  measured  from  the  given  circle,  having  as  many 
(arcual)  degrees  as  there  are  (angular)  degrees  in  the  given  angle ; 
draw  a  great  circle  through  the  extremity  of  this  arc  and  the  given 
point  of  intersection;  this  will  be  the  circle  required  (Art.  11.  c). 

(6)  To  construct  a  great  circle  passing  through  a  given  point^and 
making  a  given  angle  with  a  given  great  circle.     [When  the  given 
point  is  on  the  given  circle  this  problem  reduces  to  problem  (o).] 
It  is  easily  shown  that  the  angle  betweenxtwo  great  circles  is 
equal  to  the   angular   distance    (Art.  6.  c)  between  their  poles. 
Hence,  find  a  pole  of  the  given  circle  by  (1)  ;  about  this  point  as 
pole  describe  a  second  circle  whose  angular  radius  (Art.  6.  d)  is 
equal  to  the  given  angle ;  the  pole  of  the  required  circle  must  be 
on  this  second  circle.     About  the  given  point  as  pole  describe  a 
great  circle ;  if  the  required  problem  is  possible,  this  circle  will 
either  touch  or  intersect  the  second  circle.     The  points  of  contact 
or  intersection  are  the  poles  of  two  great  circles,  each  of  which 
will  satisfy  the  given  conditions. 

Ex.     Discuss  the  case  in  which  the  given  point  is  the  pole  of  the  given  circle. 


22  SPHERICAL    TRIGONOMETRY.  [Cn.  I. 

23.  Construction  of  triangles.      The  three  sides  and  the  three 
angles  of  a  spherical  triangle  constitute  its  six  jmrts  or  elements. 
If  any  three  of  these  six  parts  be  known,  the  triangle  can  be  con- 
structed.    The  construction  belongs  to  geometry;  the  computation 
of  the  three  remaining  parts,  when  three  parts  are  given,  is  an 
important   part   of   spherical   trigonometry.     The   sets  of  three 
parts  that  can  be  taken  from  the  six  parts  of  a  spherical  triangle 
are  as  follows : 

I.  Three  sides. 

II.  Three  angles. 

III.  Two  sides  and  their  included  angle. 

IV.  One  side  and  the  two  adjacent  angles. 

V.    Two  sides  and  the  angle  opposite  one  of  them. 
VI.   Two  angles  and  the  side  opposite  one  of  them. 

NOTE.  There  are  four  construction  problems  in  the  case  of  plane  triangles 
(Plane  Trig.,  Art.  53).  When  three  angles  of  a  spherical  triangle  are  given, 
there  is  only  one  spherical  triangle  (with  the  triangle  symmetrical  to  it),  as 
will  presently  appear,  which  satisfies  the  given  conditions.  When  three 
angles  of  a  plane  triangle  are  given,  there  is  an  infinite  number  of  triangles, 
of  the  same  shape,  but  of  different  magnitudes,  which  have  angles  equal  to 
the  three  given  angles.  Cases  IV.  and  VI.  above,  in  which  two  angles  are 
given,  reduce  to  a  single  case  in  plane  trigonometry,  namely,  the  case  in 
which  one  side  and  two  angles  are  given  ;  since  the  sum  of  the  three  angles 
of  any  plane  triangle  is  180°. 

24.  To  construct  a  spherical  triangle. 

I.   Given  the  three  sides.     On  any  great  circle  lay   off  an   arc 

equal  to  one  of  the  given  sides  [Art.  22  (3)].  About  one  extrem- 
ity of  this  arc  as  pole,  describe  a  circle  with  a  radius  (arcual) 
equal  to  the  second  of  the  given  sides ;  about  the  other  extremity 
of  the  arc  as  pole,  describe  a  circle  with  a  radius  equal  to  the 
third  of  the  given  sides.  By  arcs  of  great  circles  join  either  of 
the  points  of  intersection  of  the  last  two  circles  to  the  extremities 
of  the  arc  first  laid  off;  the  triangle  thus  formed  satisfies  the 
given  conditions. 

Ex.  1.  Compare  the  construction  in  the  corresponding  case  in  plane 
triangles. 


23-24.]  CONSTRUCTION  OF  TRIANGLES.  23 

Ex.  2.  How  many  triangles  are  possible  when  the  first  arc  is  laid  off  ? 
Are  these  triangles  equal  or  symmetrical  ? 

Ex.  3.  Construct  ABC:  (a)  Given  a  =  70°,  b  =  65°,  c  =  40°;  (6)  Given 
a  =  120°,  b  =  115°,  c  =  80°. 

Ex.  4.  Determine  approximately  the  angles  of  these  triangles.  (See 
Arts.  11.  c,  34.) 

II.  Given  the  three  angles.     Calculate   the   sides   of  the   polar 
triangle    (Art.  16.   d);   construct  it  by   I.  above;   construct   its 
polar  (Art.  16.  6)  j  the  latter  triangle  is  the  one  required. 

Ex.  1.  How  many  triangles  can  be  drawn  when  one  side  of  the  polar 
triangle  is  fixed  ?  Are  these  triangles  equal  or  symmetrical  ? 

Ex.  2.   Discuss  the  corresponding  case  in  plane  triangles. 

Ex.  3.  Construct  ABC:  (a)  Given  A  =  85°,  B  =  75°,  C  =  55° ;  (6)  Given 
A  =  75°,  B  =  105°,  C  =  100°. 

Ex.  4.   Determine  approximately  the  sides  of  these  triangles. 

III.  Given  two  sides  and  their  included  angle.     Take  any  point 
on  any  great  circle;   through  this  point  draw  a  circle  making 
with  the  first  circle  an  angle  equal  to  the  given  included  angle 
[Art.   22  (5)];   from  the  chosen  point   and  on   the  first  circle 
bounding  this  angle,  lay  off  an  arc  equal  to  one  of  the  given  sides ; 
from  the  same  point  and  on  the  second  circle  bounding  the  angle, 
lay  off  an  arc  equal  to  the  other  given  side.     Join  the  extremities 
of  these  arcs  by  the   arc  of  a   great  circle;   the  triangle  thus 
formed  is  the  one  required. 

Ex.  1.  How  many  triangles  can  be  made  when  the  first  circle  and  the 
point  are  chosen  ?  Are  these  possible  triangles  equal  or  symmetrical  ? 

Ex.  2.    Discuss  the  corresponding  case  in  plane  triangles. 

Ex.  3.  Construct  ABC :  (a)  Given  a  =  75°,  6  =  120°,  C  =  66°  ;  (6)  Given 
b  =  35°,  c  =  70°,  A  =  110°. 

Ex.  4.   Determine  approximately  the  remaining  parts  of  these  triangles. 

IV.  Given  a  side  and  its  two  adjacent  angles. 

Either :  a.  On  any  arc  of  a  great  circle  lay  off  an  arc  equal  to 
the  given  side ;  its  extremities  will  be  taken  as  two  vertices  of 
the  required  triangle.  Through  one  extremity  of  this  arc  draw  a 
great  circle  making  with  the  arc  an  angle  equal  to  one  of  the 
given  angles;  through  the  other  extremity  of  the  arc  draw  a 


24  SPHERICAL   TRIGONOMETRY.  [Cn.  I. 

great  circle  making  with  the  arc  (and  on  the  same  side  as  the 
angle  first  constructed)  an  angle  equal  to  the  other  of  the  given 
angles.  The  point  of  intersection  of  these  two  circles  which  is  on 
the  same  side  of  the  arc  as  the  two  angles,  is  the  third  vertex  of 
the  required  triangle. 

Or:  b.  Calculate  the  corresponding  parts  of  the  polar  tri- 
angle ;  construct  it  by  III. ;  construct  its  polar ;  this  will  be  the 
triangle  required. 

Ex.  1.  How  many  triangles  are  possible  when  the  first  arc  is  chosen  ? 
Are  these  triangles  equal  or  symmetrical  ? 

Ex.  2.  Discuss  the  corresponding  case  in  plane  triangles. 

Ex.  3.  Solve  Problem  III.  by  means  of  IV.  a,  and  the  polar  triangle. 

Ex.  4.  Construct  ABC :  (a)  Given  a  =  75°,  B  =  65°,  C  =  110°  ;  (6)  Given 

-6  =  110°,  A  =  40°,  C  =  63°. 

Ex.  5.  Determine  approximately  the  remaining  parts  of  these  triangles. 

V.  Given  two  sides  and  the  angle  opposite  to  one  of  them.  [First, 
review  the  corresponding  case  in  plane  geometry.] 

To  construct  a  triangle  ABC  when  a,  6,  A,  are  known :  Through 
any  point  A  of  a  great  circle  ALA' A  draw  the  semicircle,  ACA', 
making  an  angle  A  LAC  equal  to  the  given  angle  A.  From  this 
semicircle  cut  off  an  arc  AC  equal  to  b.  About  C  as  a  pole,  with 
an  arc  equal  (in  degrees)  to  the  side  a,  describe  a  circle.  The 
intersection  B  of  this  circle  with  ALA1  will  be  the  third  vertex  of 
the  required  triangle,  A  and  C  being  the  other  two  vertices. 

Four  cases  arise,  viz. :  — 

(1)  When  the  circle  described  about  C  fails  to  intersect  ALA' ; 

(2)  When  it  just  reaches  to  ALA' ; 

(3)  When  it  intersects  the  semicircle  ALA'  in  but  one  point ; 

(4)  When  it  intersects  the  semicircle  ALA>  in  two  points. 

Case  (1)  is  represented  in  Figs.  18,  22  ;  case  (2),  in  Figs.  10,  23  ;  case  (3), 
in  Figs.  20,  24  ;  and  case  (4),  in  Figs.  21,  25.  In  Figs.  18  and  22  the  angle 
A  is  respectively  acute  and  obtuse  ;  and  similarly  for  each  of  the  other  pairs 
of  figures. 

NOTE.  In  Figs.  18-25  AKA'  is  a  great  circle  in  the  plane  of  the  paper, 
and  ALA' A  is  a  great  circle  at  right  angles  to  that  plane,  ALA1  being  above 
the  surface  of  the  paper,  and  the  dotted  A  A'  being  below.  In  Figs.  18-21, 


24.] 


CONSTRUCTION  OF  TRIANGLES. 


25 


Fia.  18 


Fia.  19 


Fia.  32 


Fia.  31 


FIG.  35 


26  SPHERICAL   TRIGONOMETRY.  [Cn.  I. 

angle  A  is  acute  [equal  to  PAK(W°}-KAC~],  and  the  arc  AC  is  in  front 
of  the  paper.  In  Figs.  22-25,  angle  A  is  obtuse  [equal  to  PAK(9Q°)  -f 
1\A  C),  and  the  arc  AC  is  behind  the  paper. 

In  Fig.  21  there  are  two  triangles  (not  equal  or  symmetrical)  that  satisfy 
the  given  conditions;  amHikewisemn'Fig.  25.  Hence  V.  is  an  ambiguous 
case  in  spherical  geometry. 

In  each  figure  let  the  perpendicular  arc  CP  be  drawn  from  C  to  the  semi- 
circle ALA',  and  let  its  length  be  called^.  [See  Ex.  1,  p.  101.] 

A.  When  angle  A  is  acute : 

Fig.  18  shows  that  the  triangle  required  is  impossible,  if  CB  <  CP,  i.e.  if 
a<p. 

Fig.  19  shows  that  the  triangle  required  is  right  angled  if  CB  =  CP,  i.e. 
v   ifa=.p. 

Fig.  20  shows  that  there  is  but  one  triangle  which  satisfies  the  given  con- 
ditions, if 

CB>  CP,  CB>CA,  and  CB<  CA' } 

i.e.  if  a  >_p,  a  >  b,  and  a  <  180°  —  b. 

^  Similarly,  there  is  only  one  triangle  if  a  >p,  a<b,  and  a  >  180°  —  b. 

Fig.  21  shows  that  there  are  two  triangles  which  satisfy  the  given  condi- 
tions, if 

CB>  CP,  CB<  CA,  and  CB  <  CA' ; 

4    i.e.  if  a  >p,  a<b,  and  a  <  180°  -  b. 

B.  When  angle  A  is  obtuse : 

Fig.  22  shows  that  the  triangle  required  is  impossible,  if  CGB  >  CGP,  i.e. 
v     \ia>p. 

Fig.  23  shows  that  the  triangle  required  is  right  angled,  if  CGB  =  CGP, 
i.e.  if  a  —  p. 

Fig.  24:  shows  that  there  is  but  one  triangle  which  satisfies  the  given  con- 
ditions, if 

CLB  <  CGP,  CLB  <  CA  and  CLB  >  CA' ; 

J  i.e.  if  a  < p,  a  <  b,  and  a  >  180°  -  6. 

Similarly,  there  is  only  one  triangle  if  a  <p,  a  >  b,  and  a  <  180°  —  b. 
Fig.  25  shows  that  there  are  two  triangles  which  satisfy  the  given  condi- 
tions, if 

CLB<  CGP,  CLB>  CA,  and  CLB>  CA' ; 

V  i.e.  if  a  <p,  a  >  b,  and  a  >  180°  -  6. 

In  Fig.  25  produce  PGC  to  meet  the  great  circle  ALA  A  in  P'.     Then 
CP1  =  180° .  —  p.    Since  AC  and  CA '  are  each  greater  than  CP',  it  follows 
J  that  a >  180°  -p. 


24.]  CONSTRUCTION   OF  TRIANGLES.  27 

It  is  also  apparent  from  Figs.  20  and  21  that  the  triangle  is  impossible, 

if  A  is  acute,  a  >  &,  and  a  >  180°  -  b  ; 

and  it  is  apparent  from  Figs.  24  and  25  that  the  triangle  is  impossible, 
if  A  is  obtuse,  a  <  &,  and  a  <  180°  -  5. 

Some  special  cases  which  may  be  investigated  by  the  student, 
are  indicated  in  the  exercises  on  this  chapter  at  page  101. 

VI.  Given  two  angles  and  the  side  opposite  one  of  them.  Calculate 
the  corresponding  parts  of  the  polar  triangle ;  construct  it  by  V. ; 
construct  its  polar ;  this  is  the  required  triangle.  There  may  be 
two  solutions,  since  the  triangle  first  constructed  may  have  two 
solutions. 

Ex.  1.  Construct  ABC:  (a)  Given  a  =  52°,  6  =  71°,  A  =  46°;  (6)  Given 
a  =  99°,  6  =  64°,  A  =  95°. 

Ex.  2.  Construct  ABC  :  (a)  Given  A  =  46°,  B  =  36°,  a  =  42°  ;  (6)  Given 
A  =  36°,  B  =  46°,  a  =  42°. 

Ex.  3.    Determine  approximately  the  remaining  parts  of  these  triangles. 

N.B.  Questions  and  exercises  on  Chapter  I.  will  be  found  at  pages 
101-102. 


CHAPTER   II. 

RIGHT-ANGLED    SPHERICAL   TRIANGLES. 

25.  Spherical  trigonometry.     Spherical  trigonometry  treats   of 
the  relations  between  the  six  parts  of  a  triedral  angle,  or,  what  is 
the  same  thing  (Art.  14),  the  relations  between  the  six  parts  of  the 
corresponding  spherical  triangle  intercepted  011  the  surface  of  the 
sphere.     In  Art.  24  it  has  been  seen  how  a  triangle  can  be  con- 
structed when  any  three  parts  are  given ;   Chapters  II.  and  III. 
are  concerned  with   showing  how  the  remaining   parts   can  be 
computed  when  any  three  parts  are  known.     In  this  chapter  the 
relations  between  the  sides  and  angles  of  a  right-angled  spherical 
triangle  are  deduced.1*    Throughout  the  book  the  relations  between 
trigonometric   ratios,  discussed   in   plane  trigonometry,   will   be 
employed. 

26.  Relations  between  the  sides  and  angles  of  a  right-angled  spheri- 
cal triangle. 

Case  I.     The  sides  about  the  right  angle  both  less  than  90°. 
Let  ABC  be  a  spherical  triangle  which  is  right  angled  at  C  and 
on  a  sphere  whose  centre  is  at  0.     First  suppose  that  a  and  b  are 

B 


FIG.  26 


*  These  relations  can  also  be   obtained   from  the   relations,  derived  in 
Chapter  III.,  between  the  parts  of  any  spherical  triangle  or  triedral  angle. 

28 


25-26.]     RELATIONS  BETWEEN  SIDES  AND  ANGLES.  29 

each  less  than  90°.  (It  is  easily  shown,  geometrically,  that  c  is 
then  less  than  90°.)  Draw  OA,  OB,  OC.  Take  any  point  P  on 
OA  ;  in  the  plane  OA  C  draw  PL  at  right  angles  to  OA,  and  let 
it  meet  OC  in  L  ;  in  the  plane  OAB  draw  PM  at  right  angles  to 
OA,  and  let  it  meet  OB  in  M  ;  and  draw  ML.  Since  PL  and 
PM  are  perpendicular  to  the  line  OA,  the  line  OA  is  perpendicu- 
lar to  the  plane  LPM  (Euc.  XI.  4)  ;  and,  therefore,  the  plane 
LPM  i£  perpendicular  to  the  plane  OAC  (Euc.  XI.  18).  Also, 
the  plane  gCB  is  perpendicular  to  the  plane  OAC,  since  C  is  a 
right  angle.  Hence,  LM,  the  intersection  of  the  planes  LPM 
and  OCB,  is  perpendicular  to  the  plane  (MO  (Euc.  XI.  19)  ;  and 
hence,  MLP  and  MLO  are  right  angles. 

In  the  triangle  0PM,  the  angle  0PM  being  right, 


OM      OL     OM 

Now,  angle  POM  =  c,  ^  =  cos  POL  =  cos  6, 

OL 

and  —  =  cos  LOM  =  cos  a. 

OM 

.-.  cos  c  =  cos  a  cos  6.  (1) 

In  the  triangle  LPA/,  angle  PLM=  90°,  and  angle  LPM  =  A; 

LM 
LM      OM     sin  LOM 


OM 

(2) 


sine 


PL 

PL       OP      tan  POL 


OP 


Also, 


whence,  cos  ^  =  (3) 

tanc 


SPHERICAL    TRIGONOMETRY.  [CH.  II. 


LM 


LM      OL 

Also,  tan  LPM  =  -       =  - 


OL 

whence,  tan  A  =  (4) 

sin  6 

On  remarking  that  A,  a,  denote  an  angle  and  its  opposite,  side, 
and  that  b  denotes  the  other  side,  the  relations  for  angle  B  cor- 
responding to  (2),  (3),  (4),  can  be  written  immediately  ;  viz.  : 


(2');     cos7J  =  -  (3<);     tanS  =  (4'). 

sin  c  tan  c  sin  a 

These  relations  for  B  can  also  be  deduced  directly,  by  taking 
any  point  on  OB  and  making  a  construction  similar  to  that  shown 
in  Fig.  26. 

Moreover, 


sin  A  =  tan  A  cos  A  =  •         -  =  •  -—    [By  (3),  (4).] 

smi     tanc     tanc     cos* 


or       -i       i  •       T-»         COS  A 

Similarly.  sin  B  =  —    — •  (o) 

cos  a 

Also,  cos  c  =  cos  a  cos  b  =  ~  -^—  •  -    — •     [By  (1),  (o),  (o').] 

Sill  -Z5       S1U  -*T-L 

.\  cos  c  =  cot  A  cot  B.  (6) 

For  convenience  of  reference,  relations  (l)-(6)  may  be  grouped 
together : 

cos  c  -  cos  a  cos  b.  (1) 

elii    A  —  **tn<^.  gin  ^  —  Sin    .,  /0\     /of\ 

sin  c  sin  c 

tanf '  (3>>  (3') 

sTn^'  W,  W 

sin  B  =  55Lfl.  /5x   /5»x 

cos  a 

cos  c  =  cot  A  cot  JS.  (6) 


26.] 


RELATIONS  BETWEEN  SIDES  AND  ANGLES. 


81 


Case  II.     The  sides  about  the  right  angle  both  greater  than  90°. 
.  In  Fig.  27,  C  denotes  the  right  angle,  and  the  sides  a,  b,  are 
each  greater  than  a  quadrant. 


Form  the  lime  CO  by  producing  the  sides  a  and  b  to  meet  in 
C'..  Then  ABC'  is  a  right  triangle  in  which  the  sides  about  the 
right  angle  are  each  less  than  90°. 

.-.  cos  c  =  cos  BO  cos  AC'  =  cos  (180  -  a)  cos  (180  -  b). 


Hence 

Also, 

cos 
whence, 


tanAB 


cos  c  =^cos  aVos  b. 

cos  (180°  -  BAG)  = 
tan  b 


tan 


cos  ^4  = 


tan  c 


In  a  similar  manner  the  other  relations  in  (l)-(6)  can  be  shown 
to  be  true  for  ABC  (Fig.  27). 

NOTE.     ABC'  is  said  to  be  co-lunar  with  ABC.     Every  triangle  has  three 
co-lunar  triangles,  one  corresponding  to  each  angle. 

Case  III.     One  of  the  sides  about  the  right  angle  less  than  90°, 
and  the  other  side  greater  than  90°. 

c 


FIG.  28 


In  ACB  let  C  =  90°,  a  >  90°,  and  b  <  90°.  Complete  the  lune 
BB'.  Then  ACB'  is  a  right-angled  triangle  in  which  the  sides 
about  the  right  angle  are  each  less  than  90°. 


.32  SPHERICAL   TRIGONOMETRY.  [On.  II. 

In  ACS',  cos  AB'  =  cos  AC  cos  CB' ; 

i.e.  cos  (180°  -  c)  =  cos  b  cos  (180°  -  a) ; 

whence  cos  c  =  cos  a  cos  fr. 

Again, 

cos  CAB1  =  tan^C;  i.e.  cos  (180°  -  BAG)  = tan^C 

tan^LB''  tan  (180°  -  BA) ' 

tan  6 

whence.  cos  A  = 

tan  c 

In  a  similar  way  the  other  relations  in  (1)— (6)  can  be  shown 
to  be  true  for  ABC  (Fig.  28). 

erJV  27.  On  species.  Two  parts  of  a  spherical  triangle  are  said  to 
be  of  the  same  species  (or  of  the  same  affection)  when  both  are  less 
than  90°,  both  greater  than  90°,  or  both  equal  to  90°.  Formula 
(1),  Art.  26,  shows  that  the  kifflojgniw  of  a  right-angled  spherical 
triangle  is  less  than  90°  when  the  sides  about  the  right  angle  are 
both  greater  or  both  less  than  90° ;  and  it  shows  that  the  hypotenuse 
is  greater  than  90°  when  the  sides  are  of  different  species.  Formulas 
(4)  and  (4')  show  that  in  a  right-angled  spherical  triangle  (since 
the  sines  of  the  sides  are  positive)  an  angle  and  its  opposite  side 
are  of  the  same  species.  These  important  properties  can  also  be 
deduced  geometrically. 

EXAMPLES. 

N.  B.     It  is  advisable  to  remember  the  result  of  Ex.  1. 

1.  State  relations  (l)-(6),  Art.  26,  in  words. 

(1).    cos  hyp.    =  product  of  cosines  of  sides. 

(6).    cos  hyp.    =  product  of  cotangents  of  angles. 
(2),  (2').    sin  angle  =  sin  opposite  side  ~  sin,  hyp. 
(3),  (3').    cos  angle  =  tan  adjacent  side  -f-  tan  hyp. 
(4),  (4').    tan  angle  =  tan  opposite  side  -=-  sin  adjacent  side. 
(5),  (5').    cos  angle  =  cos  opposite  side  x  sin  other  angle. 

[Compare  (2),  (3),   (4),  with  the  corresponding  formulas  in  plane  tri- 
angles.] 

2.  Deduce  formulas  (l)-(4)  by  means  of  a  figure  in  which  P  is  anywhere 
on  OB  (see  Fig.  26). 


27-28.]    SOLUTION  OF  A   RIGHT-ANGLED   TRIANGLE.  33 

3.  Deduce  formulas  (l)-(4)  by  means  of  a  figure  in  which  P  (see  Fig.  26) 
is :  (a)  in  OA  produced ;  (6)  in  OB  produced ;  (c)  at  the  point  A  ;  (d)  at 
the  point  B. 

4.  The  two  sides  about  the  right  angle  of  a  spherical  triangle  are  60°  and 
75°  ;  find  the  hypotenuse  and  the  other  angles  by  means  of  relations  (1),  (4), 
(4'),  Art.  26.     Check  (or  test)  the  result  by  means  of  other  formulas. 

5.  In  ABC,  given  ^1  =  47°  30',  .5  =  53°  40',  C=90°;  find  the  remaining 
parts,  and  check  the  results. 

6.  Solve  some  of  the  examples  in  Art.  31,  and  check  the  results. 

28.   Solution  of  a  right-angled  triangle. 

N.B.  The  student  is  advised  to  investigate  this  subject  independently  ; 
and,  before  reading  this  article,  to  put  in  writing  in  an  orderly  manner  his 
ideas  about  the  solution  of  right  triangles.  These  ideas  will  thus  be  made 
clearer  in  his  mind,  and  his  subsequent  reading  will  be  easier. 

In  a  right  triangle  there  are  five  elements  beside  the  right 
angle.  These  five  elements  can  be  taken  in  groups  of  three  in 
ten  different  ways.  Each  of  these  ten  groups  is  involved  in  the 
ten  relations  derived  in  Art.  26;  the  three  elements  of  each 
group  are,  accordingly,  connected  by  one  relation. 

Ex.  (a)  Write  all  the  groups  of  three  that  can  be  formed  from  a,  6,  c, 
A,  B,  such  as  a,  &,  c ;  a,  b,  A ;  etc. 

(6)  Write  the  relation  connecting  the  elements  of  each  group. 

It  follows  that  if  any  two  elements  of  a  right-angled  spherical 
triangle  besides  the  right  angle  be  given,  then  the  remaining  three 
elements  can  be  determined.  The  method  of  finding  a  third 
element  is  as  follows : 

Write  the  relation  involving  the  two  given  elements  and  the  re- 
quired element;  solve  this  equation  for  the  required  element. 

Check  (or  test).  When  the  required  elements  are  obtained,  the 
results  can  be  checked  by  examining  whether  they  satisfy  rela- 
tions which  have  not  been  employed  in  the  solution,  and,  pref- 
erably, the  relation  involving  the  newly  found  elements. 

E.g.,  suppose  that  A,  &,  are  known,  C  being  90° ;  then  a,  c,  B, 
are  required.  Side  a  can  be  found  by  (4) ;  side  c,  by  (3) ;  and 
angle  B,  by  (5).  The  values  found  for  a,  c,  B,  can  be  checked 

by  (31)- 


84  SPHERICAL   TRIGONOMETRY.  [Cn.  IL 

NOTE  1.  The  cosine,  tangent,  and  co-tangent  of  sides  and  angles  greater 
than  90°,  are  negative.  Careful  attention  must  be  paid  to  the  algebraic  signs 
of  the  trigonometric  functions  appearing  in  the  work. 

NOTE  2.     The  properties  stated  in  Art.  27  are  very  useful. 

NOTE  3.  Determine  each  element  from  the  given  elements  alone.  If  an 
element  is  found  erroneously  and  then  used  in  rinding  a  second  element,  the 
second  element  will  also  be  wrong. 

The  ten  possible  groups  of  three  elements  correspond  to  the 
following  six  cases  for  solution,  in  which  the  given  elements  are 
respectively : 

(1)  Two  sides.  (4)  Two  angles. 

(2)  Hypotenuse  and  a  side.  *     (5)  Side  and  opposite  angle. 

(3)  Hypotenuse  and  an  angle.  (6)  Side  and  adjacent  angle. 

Before  proceeding  to  the  solution  of  numerical  examples,  it  is 
necessary  to  refer  more  particularly  to  one  of  these  cases ;  and 
also  to  call  attention  to  the  fact  that  the  ten  formulas  for  right 
triangles  (Art.  26)  may  be  grouped  in  two  very  simple  and  con- 
venient rules. 

j/*  29.  The  ambiguous  case.  When  the  given  parts  are  a  side  and 
its  opposite  angle,  there  are  two  triangles  which  satisfy  the  given 
conditions.  For,  in  ABC  (Fig.  29),  let  C  =  90°,  and  let  A  and  CB 
(equal  to  a)  be  the  given  parts.  Then,  on  completing  the  lune 
AA1,  it  is  evident  that  the  triangle  A'BC  also  satisfies  the  given 
conditions,  since  BOA'  =  90°,  A'  =  A,  and  CB  =  a.  The  remain- 


o  C 

Fia.  29 

ing  parts  in  A'BC  are  respectively  supplementary  to  the  remain- 
ing parts  in  ABC',  thus  A'B  =  180°-c,  A'C=18Q°-b,  A'BC 
=  180°  -  ABC. 

This  ambiguity  is  also   apparent  from  the  relations  (l)-(6), 
Art.  26.     For,  if  a,  A,  are  given,  then  the  remaining  parts,  namely, 


29-30.]  NAPIER1  S  RULES.  35 

c,  b,  B}  are  all /determined  from  their  sines  [see  (2),  (4),  (5'),] ; 
and,  accordingly,  c,  b,  B,  may  each  be  less  or  greater  than  90°. 
On  the  otMr  hand,  if,  for  instance,  a  and  c  are  given,  then  b  is 
determined  from  its  cosine  by  (1) ;  and  there  is  no  ambiguity, 
because  b  is  less  or  greater  than  90°  according  as  cos  b  is  respec- 
tively positive  or  negative. 

N.H.  Both  solutions  should  be  given  in  the  ambiguous  case,  unless  some 
information  is  given  which  serves  to  indicate  the  particular  solution  that  is 
suitable. 

30.  Napier's  rules  of  circular  parts.  The  ten  relations  derived  in 
Art.  26  are  all  included  in  two  statements,  which  are  called  Napier'' s  rules  of 
circular  parts,  after  the  man  who  first  announced  them,  Napier,  the  inventor 
of  logarithms. 

Let  ABC  be  a  triangle  right-angled  at  C.  Either  draw  a  right-angled 
triangle,  and  mark  the  sides  and  angles  as  in  Fig.  31,  or  draw  a  circle  and 
mark  successive  arcs  as  in  Fig.  32,  in  which  b,  a,  Co-B,  Co-c,  Co-A,  are 

CO-B  c°;9— -- 

/  \       /     No 


Co-c 


C  f 

\   .<-'  \    ^ 


Co-A  wv* 

Pia.  31  Fia.  33 

arranged  in  the  same  order  as  5,  a,  B,  c,  A,  in  Fig.  30.  (Here  Co-B,  Co-c, 
Co-A,  denote  the  complements  of  B,  c,  and  A,  respectively.)  The  five 
quantities,  a,  b,  Co-B,  Co-c,  Co-A,  are  known  as  circular  parts.  That  is,  the 
right  angle  being  omitted,  the  two  sides  and  the  complements  of  the  hypote- 
nuse and  the  other  ang'.es  are  called  the  circular  parts  of  the  triangle. 

In  Figs.  31  and  32  each  part  has  two  circular  parts  adjacent  to  it,  and  two 
circular  parts  opposite  to  it.  Thus,  on  taking  a,  for  instance,  the  adjacent 
parts  are  b,  Co-B,  and  the  opposite  parts  are  Co-c,  Co-A.  If  any  three  parts 
be  taken,  one  of  them  is  midway  between  the  other  two,  and  the  latter  are 
either  its  two  adjacent  parts  or  its  two  opposite  parts.  Thus,  if  a,  b,  Co-A, 
be  taken,  then  b  is  the  middle  part,  and  a,  Co-A,  are  the  adjacent  parts  ;  if  a, 
b,  Co-c,  be  taken,  then  Co-c  is  the  middle  part,  and  a,  b,  are  opposite  parts. 

Ex.  Take  each  of  the  circular  parts  in  turn,  write  its  opposite  parts  and 
adjacent  paxts,  thus  getting  ten  sets  in  all. 


T 

36  SPHERICAL    TRIGONOMETRY.  [Cn.  II 

Napier"1  s  rules  ofafacular  parts  are  as  follows  : 

I,  The  sine  of  me  middle  part  is  equal  to  the  product  of  the  tangents  oj 
the  adjacent  parts.    kr 

II.  The  sine  of  tltg  middle  part  is  equal  to  the  product  of  th'e  cosines  of 
the  opposite  parts. 

(The  I's,  a's,  and  o's  are  lettered  thus,  in  order  to  aid  the  memory.) 
These  rules  are  easily  verified.     For  example,  on  taking  a  for  the  middle 
part, 

sin  a  =  tan  b  tan  (90°  -  J3)  =  tan  b  cot  B.  [See  Art.  26  (4').] 

sin  a  =  cos (90°  -  A)  cos  (90°  -  c)  =  sin  J.,sin  c.     [See  Art.  26  (2).] 
Again,  on  taking  Co-A  for  the  middle  part, 

sin  (90°  -  A)  -  tan  6  tan  (90°  -  c),  i.e.  cos  A  =  tan  b  cot  c. 

[See  Art.  26  (3).] 
sin  (90°  —  A)=  cos  a  cos  (90°  —  _B),  i.e.  cos  A  =  cos  a  sin  B. 

[See  Art.  26  (5').] 

In  a  similar  way  each  of  the  remaining  three  parts  can  be  taken  in  turn 
for  the  middle  part,  and  the  remaining  six  relations  of  Art.  26  shown  to 
agree  with  Napier's  rules.* 

NOTE.  One  may  either  memorize  the  relations  in  Art.  26  (or  Ex.  1, 
Art.  27),  or  use  Napier's  rules.-  Opinions  differ  as  to  which  is  the  better 
thing  to  do. 

Ex.  1.  Verify  Napier's  rules  by  showing  that  they  include  the  10  relations 
in  Art.  26. 

Ex.  2.    Prove  Napier's  rules. 

31.  Numerical  problems.  In  solving  right  triangles  the  pro- 
cedure is  as  follows : 

(1)  Indicate  the  two  given  parts  and  the  three  required  parts. 

*  This  is  only  a  verification  of  Napier's  rules.  One  proof  of  the  rules 
would  consist  of  the  derivation  of  the  relations  in  Art.  26  plus  this  verifica- 
tion. These  rules  were  first  published  by  Napier  in  his  work  Mirifici 
Logarithmorum  Canonis  Descriptio  in  1614.  Napier  indicated  a  geometrical 
method  of  proof,  and  deduced  the  rules  as  special  applications  of  a  more 
general  proposition.  They  are  something  more  than  mere  technical  aids  to 
the  memory.  For  an  explanation  of  this  and  of  their  wider  geometrical 
interpretation,  see  Charles  Hutton,  Course  in  Mathematics  (edited  by  T.  S. 
Davies,  London,  1843),  Vol.  II.  pp.  24-26;  Todhunter,  Spherical  Trigo- 
nometry, Art.  68;  E.  O.  Lovett,  Note  on  Napier's  Rules  of  Circular  Parts 
(Bulletin  Amer.  Math.  Soc.,  2d  Series,  Vol.  IV.  No.  10,  July,  1898). 


31.] 


NUMERICAL   PROBLEMS. 


37 


(2)  Write  the  relations  -that  will  be  employed  in  the  solution, 
and  note  carefully  the  algebraic  signs  of  the  functions  involved. 
The  noting  of  these  signs  will  serve  to  show  (unless  a  part  is 
determined  from  its  sine)  whether  a  required  part  is  less  or  greater 
than  90°. 

(3)  For  use  as  a  check,  write  the  relation  involving  the  three 
required  parts. 

(4)  Arrange  the  work  as  neatly  and  clearly  as  possible. 

N.B.  Attention  may  be  directed  to  the  notes  in  Art.  28.  Also  see  Plane 
Trigonometry,  Art.  27  (particularly  p.  45,  notes  2,  4-6),  and  Art.  59,  p.  107. 

NOTE.  The  trigonometric  function  of  any  angle  can  be  expressed  in 
terms  of  some  trigonometric  function  of  an  angle  less  than  90°.  See  Plane 
Trigonometry,  Art.  45. 

EXAMPLES. 

1.    Solve  the  triangle  ABC,  given  : 

C  =  90°,  Solution  *  :    c  = 

a  =  44°  30',  A  = 

b  =  71°  40'  B  = 

Formulas  :  cos  c  =  cos  a  cos  b, 

tan  A  =  tan  a  -f-  sin  b, 
tan  B  =  tan  b  -^  sin  a. 


Check :    cos  c  =  cot  A  cot  B. 


Logarithmic  formulas : 
[If  necessary  ;  see  PL 
Trig.,  Art.  27,  Note  6.] 


log  sin  a  =  9. 
log  cos  a  =  9.85324  -  10 
log  tan  a  -  9.99242  -  10 
log  sin  b  =  9.97738  -  10 
log  cos  &  =  9.49768-  10 
log  tan  b  =  0.47969 

Check : 


log  cos  c  =  log  cos  a  +  log  cos  6, 
log  tan  A  =  log  tan  a  —  log  sin  6, 
log  tan  B  =  log  tan  b  —  log  sin  a, 

log  cos  c  =  log  cot  A  +  log  cot  B  (check). 
^ 
-10 


log  cos  c  =  9.35092  -  10 
logtanJ.  =  0.01504 
log  tan  B  =  0.63403 
.'.  c  =  77°    2'.1. 


=  76°  55'.  5. 


.-.  log  cot  .4  =  9. 98497  -10 
log  cot  B  =  9.36595  -  10 
.-.  log  cos  c  =  9.35092  -  10 


*  To  be  filled  in. 


••      u  - 

IE;T£ 


38 


SPHERICAL    TRIGONOMETRY. 


[Cn.  II. 


NOTE.  Spherical  triangles,  like  plane  triangles,  can  also  be  solved 
without  the  use  of  logarithms.  (See  Plane  Trigonometry,  examples  in 
Arts.  27,  55-62.) 


2.    Solve  the  triangle  ABC,  given  : 
(7=90°, 
.4  =  57°  40', 
«  =  48°30'. 


Formulas 


sine  = 


sin  B  — 


sin  a 
sin  A 
tan  a 
tan  A 
cos  A 


Solution  :  c  = 
b  = 
B  = 

Check :    sin  B  = 


sin  b 
sin  c 


cos  a 

log  sin  a  =  9.87446  -  10 
log  cos  a  =  9.82126  -  10 
log  tan  a  =  0.05319 
log  sin  A  =  9.92683  -  10 
log  cos  ^4  =  9. 72823 -10 
log  tan  A  =  0.19860 
The  check  gives  log  sin  B  =  9.90696. 


log  sin  c  =9.94763-10 
log  sin  b  =9.85459  -  10 
log  sin  B  =  9. 90697  -  10 
*    /.  c  =  62°  25'.4,  or  117°  34'.6. 
6  =  45°40'.9,  or  134°  19'.  1. 
JB  =  53°49'.3,  or  126°  10'. 7. 


On  combining  the  results   according  to  the  principles  of  Art.   27,   the 
solutions  are : 

(1)  c  =    62°25'.4,    6=    45°40'.9,    B=    53°49'.3; 

(2)  c=117°34'.6,    6  =  134°19'.l,    B  =  126°  10'. 7. 
3.    Solve  Ex.  1  without  using  logarithms. 

-       4.    Given  c  =  90°,  A  =  57°  40',  a  =  108°  30'.     Show  both  by  geometry  and 
trigonometry  why  the  solution  is  impossible. 

5.    Solve  the  triangle  ABC,  in  which  C  =  90°,  and  check  the  results,  given  : 
.     (1)    a  =    36°  25'  30",    b  =  85°  40' ;     (2)    c  =  120°  20'  30",    a  =  47°  30'  40" 'j 


v 


(3)    c  =    78°  25',  A  =  36°  42'  30" 

(5)    a  =    76°  48',  A  =  82°  36'  ; 

(7)    a  =    47°  40',  A  =  30°  43'  ; 

(9)  a  =  108°  45',  J?  =  37°42'; 
(11)  A  =  110°  27',  B  =  74°  36'  ; 
(13)  ^1  =  124°  30',  b  =  25°  40'  ; 


(4)  A  =    63°  18',          B  =  37°J' 
b  =    39°  50i  20",  A  -  47° 
P  =    TTTWSO"    B  =  80°  40*  20" 
B  =  47°  50'  ; 


(10)    c=    78°  20', 


a  =  108° 


(14)    c  =    84°  47', 


:  * 


=  39°  43'. 


7,    d  2..  rj        -/d 

32^-33.]  OBLIQUE  TRIANGLES.  39 


32.   Solution  of  isosceles  triangles  and  quadrantal  triangles. 

Isosceles  Triangles.  Plane  isosceles  triangles  can  be  solved  by 
means  of  right  triangles,  as  shown  in  Plane  Trigonometry,  Art.  32. 
A  spherical  isosceles  triangle  can  be  solved  in  a  similar  way, 
namely,  by  dividing  it  into  two  right  triangles  by  an  arc  drawn 
from  the  vertex  at  right  angles  to  the  base.  This  arc  bisects  the 
base  and  the  vertical  angle. 

Quadrantal  Triangles.  The  polar  triangle  of  a  quadrantal  tri- 
angle (Art.  18)  is  right-angled  (Art.  16.  d).  Hence,  a  quadrantal 
triangle  may  be  solved  by  solving  its  polar  triangle  by  Arts.  28, 
31,  and  then  computing  the  required  parts  of  the  quadrantal 
triangle  by  Art.  16.  d. 

EXAMPLES. 

1.  Solve  the  triangle  ABC,  in  which  A  and  B  are  equal,  and  check  the 
results,  given  : 

(1)  a  =  54°  20',  c  =  72°  54'  ;         (2)  a  =    66°  29',  A  =    50°  17'  ; 
(3)  a  =  54°  30',  (7=71°;  (4)   c  =  156°40',   (7=187°  46'. 

2.   Solve  the  trianplft..^  Kf7,  given  ; 


j>)  c  =  90°,    (7  =    67°  12',    a^EL 

(2)  c  =  90°,  A  =  136°  40',  B  =  105°  47'. 

33.*  Solution  of  oblique  spherical  triangles.  It  has  been  seen 
(Plane  Trigonometry,  Art.  34)  that  oblique  plane  triangles  can  be 
solved  by  means  of  right  triangles.  Oblique  spherical  triangles 
can  also  be  solved  by  means  of  right  spherical  triangles.  Relat- 
ing  to  spherical  triangles  there  are  six  problems  of  computation; 
these  correspond  to  the  six  problems  of  construction  discussed  in 
Arts.  23,  24.  If  any  three  parts  of  a  triangle  are  given,  the  tri- 
angle can  be  constructed  and  the  remaining  parts  can  be  com- 
puted. The  several  cases  for  computation  will  now  be  solved 
with  the  help  of  right-angled  triangles,  f 

(In  the  figures  in  this  article  the  given  parts  are  indicated  by  crosses.) 

N.B.  The  student  is  advised  to  try  to  solve  Cases  II.-VI.  before  reading 
the  text. 

*  When  time  is  limited  this  article  may  be  omitted,  or  merely  glanced  over. 
t  Other  methods  of  solving  triangles  are  shown  in  Chap.  IV. 


40 


SPHERICAL    TRIGONOMETRY. 


[Cn.  II. 


Case  I.    Given  the  three  sides.     In  ABC  (Figs.  33,  34)  let  a,  6,  c, 

be  given,  and  A,  B,  C,  be  required.  From  C  draw  CD  at  right 
angles  to  AB,  or  AB  produced.  Let  the  segments  AD  and  DB  be 
denoted  by  m  and  n,  respectively.  If  the  direction  from  A  to  B 
is  taken  as  the  positive  direction  along  the  arc  AB,  then  m  is 
positive  in  Fig.  33  and  negative  in  Fig.  34,  while  n  is  positive  in 
both  figures. 

C  C 


FIG.  33  Fio.  34 

Special  formula.     In  each  figure, 

cos  a  —  cos  n  cosp,  and  cos  b  =  cos  m  cos  p. 
cos  a      cos  b 


cosp  = 


cos  n     cos  m 


.    cosn  _cosa 
cosm     cos  b 

cos  n  —  cos  m     cos  a  —  cos 


[Composition  and  division.] 


cos  n  +  cos  m     cos  a  +  cos  b 
From  this,  on  applying  Plane  Trigonometry,  Art.  52,  formulas  (7),  (8), 
tan  i  (n  +  m)  tan  |  (n  -  m)  =  tan  |  (a  +  6)  tan  |  (a  -  6).        (1) 


Now  n  +  m  =  c]  hence,  n  —  m  can  be  found  by  (1).  Then  the 
segments  m  and  n  can  each  be  determined.  The  triangles  ADC 
and  jBZ><7  can  then  be  solved  by  Arts.  28,  31  ;  and  the  solution  of 
ABC  can  be  obtained  therefrom. 

Ex.  1.    Solve  Exs.  1,  2,  Art.  42,  by  the  method  outlined  above. 

Ex.  2.  Show  how  to  solve  this  case  when  the  perpendicular  is  drawn 
from  A  to  BC. 


33.] 


OBLIQUE  TRIANGLES. 


41 


Case  II.  Given  the  three  angles.  Solve  the  polar  triangle  by 
the  method  used  in  Case  I. ;  and  therefrom  (Art.  16.  d)  compute 
the  parts  of  the  original  triangle. 

Ex.    Solve  Exs.  1,  2,  Art.  43,  by  this  method. 

Case  III.    Given  two  sides  and  their  included  angle.      In   ABC 

(Fig.  35)  let  a,  c,  B,  be  given.     Draw  AD  at  right  angles  to  BC, 

or  BC  produced. 

In  ABD,  c  and  B  are  known ;  hence, 
BAD}  AD,  and  BD  can  be  found.  In 
ADC,  AD  and  DC  (equal  to  a  —  BD) 
are  now  known ;  hence  DAC,  ACD,  and 
AC  can  be  found.  Also,  CAB  =  CAD 
+  DAB.  The  student  can  examine  the 
case  in  which  AD  falls  outside  ABC. 

Ex.  1.    Show  how  to  solve  the  triangle  by 
drawing  a  perpendicular  arc  from  C  to  AB. 
Ex.  2.   Solve  Exs.  1,  2,  Art.  44,  by  means 


of  right  triangles. 
y         — -"^ 


Two  meth- 


Case  IT.    Given  a  side  and  the  two  adjacent  angles. 

ods  of  solution  may  be  employed. 

Either:  (1)  Solve  the  polar  triangle  by  the  method  used  in 
Case  III. ;  and  therefrom  compute  the  parts  of  the  original 
triangle. 

Or:  (2)  In  ABC  (Fig.  36)  let  A,  B,  c,  be  given.  Draw  the  arc 
BD  at  right  angles  to  AC.  In  ADB, 
AD,  DB,  and  ABD  can  be  found,  since 
A  and  AB  are  known.  Now  DBC  = 
ABC -ABD.  In  DBC,  DB  and  DBC 
are  now  known ;  hence"  BC,  CD,  and  C 
can  be  found.  Then  AC  =  AD  +  DC. 

The  student  can  examine  the  case  in 
which  BD  falls  outside  ABC. 

Ex.  1.    Solve  the  triangle  by  drawing  a  dif- 
ferent perpendicular. 

Ex.  2.   How  may  solution  (2)  aid  in  the  solution  of  Case  III.  ? 
Ex.  3.  Solve  Exs.  1,  2,  Art.  45,  by  means  of  right  triangles. 


42 


SPHERICAL    TRIGONOMETRY. 


[Cn.  II. 


Case  V.  Given  two  sides  and  the  angle  opposite  one  of  them.     (Tliis 
may  be  an  ambiguous  case;  see  Art.  24,  V.) 

In  ABC  (Fig.  37)  let  a,  c,  A,  be  given.     From  B  draw  the  arc 
BD  at  right  angles  to  AC  to  meet  AC  or 
AC  produced.      In  ABD,   c   and    A    are 
known ;  hence  AD,  DB,  and  ABD  can  be 
found.      In    DBC,    DB    and    a    are    now 
known;    hence   DBG,   C,  and  DC  can  be 
found.     Then  AC  =  AD  +  DC,  and 
=  ABD  + 


Ex.  1.   Examine  the   cases  in  which  BD  falls 
outside  ABC. 

Ex.  2.   Examine  the  case  in  which  two  triangles  satisfy  the  given  con- 
ditions. 

Ex.  3.   Solve  Exs.  1,  2,  Art.  46,  by  means  of  right  triangles. 

Case   VI.     Given  two  angles  and  the  side  opposite  one  of  them. 

Like   Case  V.  this   may  be  ambiguous;    see  Art.  24,  VI.     Two 

methods  of  solution  may  be  employed. 

Either:    (1)   Solve  the  polar  triangle  by  the  method  used  in 

Case  V. ;  and  therefrom  compute  the  parts  of  the  original  triangle. 
Or:   (2)  In  ABC  (Fig.  38)  let  A,  C,  c,  be  known.     From  B 

draw  BD  at  right  angles  to  AC  to  meet  AC,  or  AC  pro- 
duced. Solve  the  triangle  ABD;  then 
solve  the  triangle  DBO.  The  parts  of 
ABC  can  be  computed  from  these  solu- 
tions. 

Ex.  1.   How  may  (2)  aid  in  the  solution  of 
Case  V.  ? 

Ex.  2.   Solve  Exs.  1,  2,  Art.  47,  by  means  of 
right  triangles. 

Ex.  3.   Solve    the    numerical    examples    in 
Art.  24. 

34.   Graphical  solution  of  (oblique  and  right)  spherical  triangles. 

A  plane  triangle  can  be  solved  graphically  by  drawing  to  scale 
a  triangle  that  satisfies  the  given  conditions,  and  then  measuring 
the  required  parts  directly  from  the  figure  (Plane  Trigonometry, 


34.]  GRAPHICAL   SOLUTION.  48 

Arts.  10,  21).  A  spherical  triangle  can  be  solved  graphically  by 
drawing  (Art.  24)  upon  any  sphere  a  triangle  that  satisfies  the 
given  conditions,  and  then  measuring  the  required  parts  of  the 
triangle.  The  sides  and  angles  (see  Art.  11.  c)  can  be  measured 
with  a  thin,  flexible,  brass  ruler,  on  which  a  length  equal  to  a 
quadrant  or  a  semicircle  of  the  sphere  is  graduated  from  0°  to 
90°  or  180°  respectively. 

Small  slated  globes  can  be  obtained  fitting  into  hemispherical  cups,  whose 
rims  are  graduated  from  0°  to  180°  in  both  directions.  With  such  a  globe, 
cup,  and  a  pair  of  compasses,  the  constructions  discussed  in  Art.  24  and  the 
measurements  referred  to  in  this  article  are  easily  made. 

If  the  student  has  the  means  at  hand,  it  is  advisable  for  him  to 
solve  some  of  the  numerical  problems  graphically. 

N.B.     Questions  and  exercises  on  Chapter  II.  will  be  found  at  p.  102. 


CHAPTER    III. 

RELATIONS    BETWEEN   THE    SIDES  AND   ANGLES    OF 

SPHERICAL    TRIANGLES. 

i. 

35.  In   this'  chapter   some   relations   between   the   sides   and 
angles  of  any  spherical  triangle  (whether  right-angled  or  oblique) 
will  be  derived.     In  the  next  chapter  these  relations  will  be  used 
in  the  solution  of  practical  numerical  problems.     The  first  two 
general   relations    (namely,   the   Law  of  Sines  and  the  Law  of 
Cosines),  which  are  by  far  the  most  important,  can  be  derived  in 
various  ways.     In  a  short  course  it  may  be  best  to  deduce  these 
laws  by  means  of  the  properties  of  right-angled  triangles  as  set 
forth  in  Art.  26;  and,  accordingly,  this  method  is  adopted  here. 
These  laws  are  also  derived  directly  from  geometry  in  Note  A  at 
the  end  of  the  book.     It  may  be  stated  here  that  the  geometrical 
derivation  will   strengthen  the    student's  understanding   of   the 
subject,  and  will  show  more  clearly  the  correspondence  (Art.  14) 
between   the   parts  of   a  spherical  triangle  and  the  parts  of  a 
triedral  angle. 

36.  The  Law  of  Sines  and  the  Law  of  Cosines  deduced  by  means 
of  the  relations  of  right-angled  triangles. 

A.   Derivation  of  the  Law  of  Sines. 

Let  ABC  (Figs.  39,  40)  be  any  spherical  triangle.  From  B 
draw  the  arc  BD  at  right  angles  to  AC  to  meet  AC,  or  AC  pro- 
duced, in  D. 

In  ABD,          sin  BD  —  sin  c  sin  A ; 
in  CBD,  sin  BD  =  sin  a.  sin  C  (Fig.  39) 

=  sin  a  sin  BCD  (Fig.  40)  =  sin  a  sin  BCA. 
Hence,  in  both  figures,    sin  a  sin  C  =  sin  c  sin  A. 

sin  a  __  sin  c 
sin  A     sin  C 
44 


35-36.] 


LAW  OF  COSINES. 


45 


Similarly,  by  drawing  an  arc  from  C  at  right  angles  to  AB,  it 
sin  a       sin  b 


oan  be  shown  that 


sin  A      sin  B 

.    sin  a  _  sin  b 
sin  A     sinJ5 


sine 
sin  C 


V 


(1) 


In  words :  In  a  spherical  triangle  the  sines  of  the  sides  are  pro- 
portional to  the  sines  of  the  opposite  angles.  (Compare  Plane 
Trigonometry,  Art.  54,  I.) 


B.  Derivation  of  the  Law  of  Cosines. 

cos  BC  =  cos  CD  cos  DB 

=  cos '(6  -  AD)  cos  DB,  or  cos  (AD  -  b)  cos  DB 
=  cos  b  cos  AD  cos  DB  -f  sin  b  sin  J.Z)  cos  DB. 

But  f>na   Xl  /•)  r>r»«  D72  —  nne  /^  •        ft  A- -J  ,  $/2>r 


(a) 


and 


cos  ^4£>  cos  DB  =  cos  c ;     <txr< 
sin .4Z)  cos  D£  =  £2iCjm^D  =  cos c  tan ^ 


IX 


(2) 


=  cos  c  tan  ccosA=  sin  c  cos  A. '  ^ 

Hence,  on  substituting  in  (a), 

cos  a  =  cos  b  cos  c  -f-  sin  b  sin  c  cos  A . 

Similarly,  or  by  taking  the  sides  in  turn, 

cos  b  =  cos  c  cos  a  +  sin  c  sin  a  cos  B, 
cos  c  —  cos  a  cos  b  +  sin  a  sin  b  cos  C. 

In  words :  In  a  spherical  triangle  the  cosine  of  any  side  is  equal 
to  the  product  of  the  cosines  of  the  other  two  sides  plus  the  product  of 
the  sines  of  these  two  sides  and  the  cosine  of  their  included  angle. 
(Compare  Plane  Trigonometry,  Art.  54,  II.) 


46  SPHERICAL   TRIGONOMETRY.  [Cn.  III. 

NOTE  i.  The  law  of  cosines,  (2),  is  the  fundamental  and  the  most 
important  relation  in  spherical  trigonometry.  For,  as  shown  in  Note  A, 
it  can  be  deduced  directly  ;  the  law  of  sines,  (1),  can  be  deduced  from  it ; 
all  other  relations  follow  from  these  ;  and  the  relations  for  right  triangles, 
Art.  26,  can  be  deduced  from  the  relations  for  triangles  in  general,  on  letting 
C  be  a  right  angle.  The  formulas  for  cos  a,  cos  6,  cos  c,  were  known  to  the 
Arabian  astronomer  Al  Battani  in  the  ninth  century.  (See  Plane  Trigo- 
nometry, p.  166.) 

C.  The  Law  of  Cosines  for  the  angles.  Relation  (2)  holds  for 
all  triangles,  and,  accordingly,  for  A'B'C',  the  polar  triangle  qf 
ABC.  (See  Fig.  14,  Art.  16.)  That  is, 

cos  a'  =  cos  b'  cos  c'  +  sin  b'  sin  c'  cos  A9. 
.-.  cos  (180°  -^A)  =  cos  (180°  -  B)  cos  (180°  -  C) 
+  sin  (180°  -  J3)  sin  (180°  -  C)  cos  (180°  -  a).     [Art.  16.  d.] 
.*.  —  cos  A.  =  (—  cos  B)  (—  cos  (7)  -f-  sin  J5  sin  (7(—  cos  a). 

.*.  cos  A  =  -  cos  B  cos  C  +  sin  B  sin  C  cos  a.    \/  (3) 

Similarly,  cos  B  =  —  cos  C  cos  A  -f-  sin  C  sin  A  cos  6, 
cos  C  =  —  cos  A  cos  B  -f  sin  A  sin  B  cos  c. 

Relation  (3)  can  also  be  derived  by  means  of  right-angled  tri- 
angles. 

NOTE  2.     From  (2),       cos  A  =  COS  a  ~  COS  b  COS  C. 
sin  6  sin  c 

h     The  denominator  in  the  second  member  is  always  positive.     If  ^differs 
I  more  from  90°  than  does  6,  then  cos  a  is  numerically  greater  than  cos  6,  and, 
/accordingly,  greater  than  cos  b  cose;  hence  cos  A  and  cos  a  have  the  same 
/  sign,  and  thus,  a  and  A  are  in  the  same  quadrant. 

Similarly,  a  and  A  are  in  the  same  quadrant  when  a  differs  more  from 
V^gO^than  does  c. 

~~From  (3),  hTwhich      cos  a  =  cos  A  +  cos  B  cos  C, 

sin  B  sin  C 

it  can  be  shown  in  a  similar  way  that  if  A  differs  more  from  90°  than  does 
B  or  (7,  then  a  and  A  are  in  the  same  quadrant. 

Ex.  1.    Derive  cos  6  and  cos  c  by  means  of  right  triangles. 
Ex.  2.    Derive  cos  A  and  cos  B  by  means  of  right  triangles. 
Ex.  3.   Derive  cos  C  from  cos  c  by  means  of  the  polar  triangle. 


37.]        tffc  THE  HALF-ANGLES.  47 

37.   Formulas  for  the  half-angles  and  the  half-^Hfc 

[Compare  the  method  and  results  of  this  article  with  those  of 
Art.  62,  Plane  Trigonometry']. 

I.  The  half-angles. 

-™          A   j    o/»   /o\  A      cos  a  —  cos  b  cos  c  /-,  N 

From  Art.  36.  (2),  cos  A  =  —  —  (1) 

sin  b  sin  c 

cos  a  —  cos  b  cos  c 

/.  1  —  cos  A  =  L 

sin  b  sin  c 

_  cos  b  cos  c  +  sin  b  sin  c  —  cos  a 
sin  6  sin  c 

_  cos  (6  —  c)  —  cos  a 
sin  6  sin  c 

.    o  sin2 i  ^  =  2  sin  J-  (a  —  &  -f  c)  sin  j-  (a  +  6  —  c)  t 

sin  6  sin  c 

[Ptae  Trigonometry,  Art.  52,  (8).] 
On  putting  a  +  b-\-c  =  2s,  then  —  a  +  b  -f-  c  =  2  (s  —  a), 

a  —  b  +  c  =  2  (s  —  fr),  and  a  +  b  —  c  =  2  (s  —  c). 


sin2 1  A  —  s^n  (g  ~  ^  s*n  (8  ~  CX 
sin  b  sin  c 

Similarly,  from  (1), 

l  +  cos^^l  +  cosffi-cos6cosc 
sin  &  sin  c 


'. 


_  cos  a  +  sin  b  sin  c  —  cos  b  cos  c 
sin  V6  sin  c 

_  cos  a  —  cos  (6  +  c)_ 
sin  b  sin  c 


/.  2  cos2  !  ^1  =  2  sin 

sin  6  sin  c 

.'.  cos2  J  ^  =  gin*  sin  (^  -a).  (3) 

sin  6  sin  c 


48 


SPHERIC  A  L   TRIGONOMETE  Y. 


[Cu.  III. 


and  hence,          tan  *  A  =  J»in(* -»)  sin  (*-<£. 

*      sin  s  sm  (s  —  a) 


(4) 


Therefore,     tan^  = * /sin(«-a)sin(8-6)siD^--c)t 

sin(s— a)   *  sins 


hence,  if  tan  r  =  ->m  (•  -  «)  sin  (8  -  6)  sin  (g  -  c) , 

sin  s 

then  tan  \  A  = 


Like 

written 
namely, 


formulas  can  be  similarly  derived  for  J  J5  and  £  (7;  or  they  may  be 
immediately  on  observing  the  symmetry  in  formulas  (4)  and  (5); 


_    / 

' 


tan  A  B  = 


sin  a  sin  c 

sing  sin(g  —  ft) 
sin  a  sin  c 

in  (g  —  a)  sin  (g— c) 
sin  g  sin  (g— ft) 

tanr 


=A/ 


COS  £ 


sin(g  — a)sin(g— ft) 
sin  a  sin  ft 


=    /sing  sin (g  -  c) 


sin  a  sin  b 


tan  }  (7  --JsinQ-a)sinQ- 


sin  s  sin  (s  —  c) 
tanr 


sin  (s- ft) 


sin  (s  —  c) 


It  is  shown  in  Art.  50  that  r  is  the  radius  of  the  circle  inscribed 
in  the  triangle  ABC.  Article  50  may  be  read  at  this  time. 

NOTE.  By  geometry,  2  g  <  360°  and  ft  +  c  >  a.  Hence,  g  —  a  is  positive 
and  less  than  180°.  Similarly,  g  —  6,  g  —  c,  are  positive.  Therefore,  the 
quantities  under  the  radical  signs  are  positive.  The  positive  signs  must  be 
given  to  each  radical,  for  ^4,  B,  C,  are  each  less  than  180°,  and,  consequently, 
IA,1B,IC,  are  each  between  0°  and  90°. 


EXAMPLES. 

1.  Derive  each  of  the  above  formulas. 

2.  Given  a  =  58°,  b  =  80°,  c  =  96°.     Find  A,  B,  C. 

3.  Given  a  =  46°  30',  b  =  62°  40',  c  =  83°  20-.     Find  A,  B,  C. 
The  results  in  Exs.  2,  3,  may  be  checked  by  Art.  36,  (1). 


37.]  THE  HALF-SIDES.  49 

II.  The  half-sides.     From  Art.  36,  (3), 

cos  ^  +  cos  -B  CQS  ff 


cos  a  — 


sin  5  sin  (7 

On  finding  1  —  cos  a  and  1  +  cos  a,  combining  and  simplifying 
in  the  manner  followed  for  the  half-angles,  and  putting 


the  following  formulas  are  obtained : 
sin  J  a  = 


cos    a  = 
" 


tan  --  a  = 
2 


Let 


'cos(S- 
an  l\  cos 

Similarly,  or  from  (8)  and  (9)  by  symmetry, 


then  tan  |  a  -  tan  R  cos (8  — A). 


(8) 


(9) 


sin  A  sin  C  '        sin  A 


sin  ^1  sin  C  '  sin  A  sin  I? 


Ian  a  &  -A008  #  CM  (0-.g  anir    -A     -<*» 

-  - 


tan  I  6  -  tan  U  cos  (5f  -  B)  ,        tan  |  c  =  tan  JB  cos  («  -  C)  .  (11) 

It  is  shown  in  Art.  49  that  R  is  the  radius  of  the  circumscribing 
circle  of  the  triangle  ABC.     Article  49  may  be  read  at  this  time. 

NOTE  1.     Formulas  (8)-(ll)  can  also  be  derived  from  formulas  (4)-(7) 
by  making  use  of  the  polar  triangle,  as  done  in  Art.  36.  C. 


Xl^oxE  2.     Since  A+B+C  lies  between  180°  and  540°  (Art.  17),  S  lies 
/between  90°  and  270°  ;  hence,  cos  S  is  negative,  and,  accordingly,  —  cos  S 
I  is  positive.     Since  all  the  other  functions  under  the  radical  signs  are  positive, 
V^e  whole  expression  under  each  radical  sign  is  positive. 

NOTE  3.     The  positive  value  of  the  radical  is  taken,  since  each  side 
(Art.  19)  is  less  than  180° 


50  SPHERICAL   TRIGONOMETRY.  [Cu.  III. 

EXAMPLES. 

1.  Derive  formulas  (10)  from  the  values  of  cos  b  and  cose. 

2.  Derive  formulas  (10)  from  formulas  (6)  by  means  of  the  polar  trial 

3.  In  ABC,  given  A  =  78°  40',  B  =  63°  50',  C  =  46°  20'.     Fin 
[SUGGESTION.     Either  use  formulas  (8)-(10);  or,  solve  the  polar  triangle, 

and  thence  obtain  the  parts  of  the  original  triangle.     [The  results  may  be 
checked  by  using  both  these  methods,  or  by  Art.  36,  (1).] 

4.  In  ABC,  given  A  =  121°,  B  =  102°,  C  =  68°.     Find  a,  b,  c. 

5.  Show  that  cos  (S  —  A)  is  positive. 

38.  Napier's  Analogies.    On  dividing  tan  |  A  by  tan  |  B  (Art.  37), 

there  is  obtained, 

tan  i  A  _  sin  (s  —  6) 

tan  i  B  ~~  sin  (s  —  a) 
From  this,  by  composition  and  division, 

tan  A  A  -f  tan  |  B  _  sin  (s  —  b)  +  sin  (s  —  a) 
tan  \A  —  tan  1  B  ~  sin  (s  —  b)  —  sin  (s  —  a)  * 

This,  by  Plane  Trigonometry,  Arts.  44.  B,  52  (also,  see  Art.  61), 
reduces  to 

sin  i  A  cos  i  J3  +  cos  i  ^  sin  ^  ^  _  2  sin  1  (2  s  —  a  —  b)  cos  i  (a  —  b)  ^ 
sin  ±Acos±B  —  cos  ^  J.  sin  1  J3  ~~  2  cos  %  (2  s  —  a  —  6)  sin  1  (a  —  6)  ' 

sin^  +  B)          tan|c 

and  thence,  to         —  ?  -  =  -  ?  --  (i) 

sinl(^-B)     tani(a-6) 

On  multiplying  tan  ^  J.  by  tan  |  -B,  there  is  obtained 


sn  s 
i  e  sin  ±  A  sin  1  1?  _  sin  (8  —  c) 

cos  ^  A  cos  |  ^          sin  s 
From  this,  by  composition  and  division, 
cos  \  A  cos  ^  B  —  sin  i  ^4  sin  i  B      sin  s  —  sin  (s  —  c) 
cos  ±Acos±B+  sin  ±Asin^B~  sin  s  +  sin  (s  —  c) 

2  cos  |-  (2  s  —  c)  sin  i  c 
~~  2  sin  i  (2  s  —  c)  cos  £  c 

cosJ-Oi  +  B)          tan|c 
Whence,  -  =  -  ?  -  .  (2) 


38.]  NAPIER'S  ANALOGIES.  51 

Either,  on  proceeding  in  a  similar  way  with  tan  \  a  and  tan  \  b 
[Art.  37,  (8),  (10)],  or,  on  applying  (1)  and  (2)  to  the  polar  tri- 
angle, there  is  obtained, 

sin  -  (a  +  6)  cot  £  C 

(3) 
sin  *  (a -6)     tan  |  (^1-1?) 

cosi(a  +  6)  cot|C 

and  -  =  — p-       —  (4) 

cos  I  (a  —  0)     tan  |  (.4  +  1?) 

Relations  (l)-(4)  are  known  as  Napier 's  Analogies.* 

NQTE  1.     Compare  (3)  with  formula  (2)   Art.  61,  Plane  Trigonometry. 
/NOTE  2.     The  numerators  in  (3)  are  always  positive,  since  a+  6  +  c  <  360° 
y&nd  C<  180°.     It  follows,  accordingly,  that  a  -  b  and  A-  B  must  have  the 
I  same  sign.     This  also  follows  from  the  geometrical  fact  [Art.  15,  (5)]  that 
/  the  greater  angle  is  opposite  the  greater  side. 

NOTE  3.  In  relation  (4),  cot  \  C  and  cos  \(a  —  6)  are  positive  quantities  ; 
/  hence  cos£(a  +  6)  and  tan  %(A  +  B}  have  the  same  sign  ;  and,  accordingly, 
X^&(g^+  6)  and  %(A  -f  jB)  are  of  the  same  species  (Art.  27). 

NOTE  4.  Derivation  of  (3)  by  applying  (1)  to  the  polar  triangle.  On 
applying  (1)  to  the  polar  triangle  A'B'C'  (Fig.  14,  Art.  16), 

sin  \  (A*  +  B')  _        tan  ^  c' 
sin  \(A'  -  B'}  ~  tan  £(«'  -  b')' 


tan  £(180°  -  C) 

^ 


sin  $(180°  -  a  -  180°  -  6)      tan  $(180°  -  A  -  180°  -  B)  ' 
-  £  a  +  fr)  _  tan(90°  -  £  O) 


Whence 


sin  J(6  -  a)  tan  $(B  -  A) 

5)  _ 


sin  £(«-&)      tan  i  (^1-5) 

NOTE  5.  For  a  geometrical  deduction  of  Napier's  Analogies  and  the  for- 
mulas in  Art.  39,  see  M'Clelland  and  Preston,  Treatise  on  Spherical  Trig- 
onometry, Part  L,  Art.  56,  and  the  article  Trigonometry  in  the  Encyclopedia 
Britannica  (9th  edition). 

*  That  is,  Napier's  proportions.  For  a  long  time  the  word  analogy  was 
used  in  English  in  one  of  its  original  Greek  meanings,  namely,  a  proportion 
(i.e.  an  equality  of  ratios).  This  use  of  the  word  is  now  obsolete,  and  is 
only  retained  in  a  few  phrases  such  as  the  above.  Napier  (see  Art.  30,  and 
Plane  Trigonometry,  Art.  1)  discovered  these  proportions  and  gave  them 
in  his  work,  Mirifici  logarithmorum  canonis  descriptio,  in  1614 


52  SPHERICAL   TRIGONOMETRY.  [Cn.  III. 

EXAMPLES. 

1.  Express  Napier's  Analogies  in  words. 

2.  Write  the  analogies  involving  B  and  C,  A  and  C,  b  and  c,  a  and  c. 

3.  Derive  some  of  the  analogies  in  Ex.  2. 


).   Delambre's  Analogies  or  Gauss's  Formulas. 
/*        By  Plane  Trigonometry,  Art.  46,  (1), 

sin  %(A  +  .B)  =  sin  £  A  cos  £  1?  +  cos  £  J.  sin  £ 
By  Art.  37,  (4),  (6),  -         


^-8ipO»-*OcoslC, 


sin  a  sin  6  sin  c 


and        p.™  j  ^  ««  i  p  -  sin  (g  ~  «)      sin  ^  sin  (*  ~  «)  -  sm  (s  -  a)  CQC     ^ 
sin  c        '      sin  a  sin  6  sin  c 


sin  c 
«-o-6)cos|(o-6) 


In  a  similar  way  it  may  be  shown  that 

!,)  =  «»*£  -»«»IC,  (2) 


B)  =  sinja  (4) 

Formulas  (l)-(4)  are  known  as  Delambre's  Analogies,  and  also  as  Gauss's 
Formulas  or  Equations.* 


*  These  formulas  were  discovered  by  Karl  Friedrich  Gauss  (1777-1855), 
one  of  the  greatest  of  German  mathematicians  and  astronomers,  and  pub- 
lished without  proof  in  his  Theoria  Motus  Corporum  Ccelestium  in  1809 ; 
thus  they  bear  his  name.  They  were,  however,  published  earlier  by  Karl 
Brandon  Mollweide  of  Leipzig  (1774-1825)  in  Zach's  Monatliche  Correspon- 
dcnz  for  November,  1808.  They  were  earliest  discovered  by  Jean  Baptiste 
Joseph  Delambre  (1749-1822),  a  great  French  astronomer,  in  1807,  and  pub- 
lished in  the  Connaissance  des  Temps  in  1808.  The  geometrical  proof  (see 
Note  5,  Art.  38)  was  the  one  originally  given  by  Delambre.  This  proof  was 
rediscovered  and  announced  by  M.  W.  Crofton  in  1869,  and  published  in  the 
Proceedings  of  the  London  Math.  Soc.,  Vol.  III.  (1869-1871),  p.  13. 


39-40.]  DELAMBRE'S  ANALOGIES.  53 

NOTE  1.  Equations  (3)  and  (4)  can  also  be  derived  by  applying  (1)  and 
(2)  to  the  polar  triangle.  >-«VT 

NOTE  2.  Delarabre's  Analogies  can  also  be  deduced  by  help  of  Napier's 
Analogies.  (See  Todhunter,  Spherical  Trigonometry,  Art.  54  ;  Nature,  Vol. 
XL.  (1889,  Oct.  31),  p.  644.) 

NOTE  3.  On  the  other  hand,  Napier's  Analogies  can  be  easily  derived 
from  Delambre's  Analogies ;  namely,  on  dividing  corresponding  members, 
one  by  the  other,  in  (1)  and  (3),  (2)  and  (4),  (4)  and  (3),  (2)  and  (1). 

EXAMPLES, 

1.  Write  Delambre's  Analogies  involving  B  and  C,  and  G  and  A. 

2.  Derive  (3)  and  (4)  from  (1)  and  (2),  using  the  polar  triangle. 

3.  Derive  Delambre's  Analogies  from  Napier's  Analogies. 

4.  Derive  some  of  the  analogies  in  Ex.  1  directly. 

40.   Other  relations  between  the  parts  of  a  spherical  triangle.     The 

preceding  articles  of  this  Chapter  present  few  more  relations  than 
are  required  for  the  solution  of  spherical  triangles.  Between  the 
parts  of  a  spherical  triangle  there  are  many  other  relations  which 
are  interesting  and  useful  for  many  purposes,  and  which  either 
set  forth,  or  lead  to  the  discovery  -of,  important  geometrical  prop- 
erties *  of  spherical  triangles.  t 

For  example,  if  in  equation  (2)  Art.  36,  the  value  of  cos  c  in  the  second 
equation  that  follows,  be  substituted,  then 

cos  a  =  cos  a  cos2  b  +  sin  a  sin  b-  cos  b  cos  C  -f  sin  b  sin  c  cos  A  ; 
•whence,  on  putting  for  cos2  b   its  value  1  —  sin2  6,  dividing  by  sin  6,  and 
transposing,  it  follows  that 

cos  a  sin  b  —  sin  a  cos  b  cos  C  =  sin  c  cos  A. 

Eive  similar  relations  can  be  derived  by  permuting  the  letters ;  and  on 
applying  these  six  relations  to  the  polar  triangle,  six  others  can  be  derived. 

To  pursue  this  topic  further  is  beyond  the  scope  of  this  book, 
which  aims  to  give  little  more  than  the  simplest  elements  of 
spherical  trigonometry  and  what  is  absolutely  required  for  the 
solution  of  spherical  triangles.  Those  who  are  interested  can 
refer  to  the  works  on.  spherical  trigonometry  by  M'Clelland  and 
Preston  (Macmillan  &  Co.),  Casey  (Longmans,  Green,  &  Co.), 
Bowser  (D.  C.  Heath  &  Co.),  and  others. 

N.B.     Questions  and  exercises  on  Chapter  III.  will  be  found  on  page  104. 

* — , 

*  Instances  in  which  geometrical  properties  are  deduced  by  means  of 
trigonometry,  are  given  in  Art.  27,  Art.  36,  (Note  2),  Art.  38,  (Notes  2,  3). 


CHAPTER   IV. 

SOLUTION   OF  TRIANGLES. 

N.B.  The  student  is  recommended  to  work  one  or  two  examples  in  each 
set  in  this  chapter  before  reading  any  of  the  text. 

41.  Cases  for  solution.  This  chapter  is  concerned  with  the 
numerical  solution  of  spherical  triangles.  In  all  there  are  six 
cases  for  solution ;  these  correspond  respectively  to  the  six  cases 
for  construction  which  were  discussed  in  Arts.  23,  24.  In  these 
cases  the  given  parts  are  as  follows : 

I.  Three  sides. 

II.  Three  angles. 

III.  Two  sides  and  their  included  angle. 

IV.  One  side  and  its  two  adjacent  angles. 

V.    Two  sides  and  the  angle  opposite  one  of  them. 
VI.    Two  angles  and  the  side  opposite  one  of  them. 

With  slight  changes  the  procedure  described  in  Art.  31  is  rec- 
ommended. Figures  may  be  helpful.  Of  formulas  adapted  for 
logarithmic  computation,  the  necessary  ones  are  (1)  Art.  36,  (4)- 
(11)  Art.  37,  and  (l)-(4)  Art.  38.  If  the  polar  triangle  is  used 
in  finding  the  solution,  then  I.  and  II.  constitute  one  case,  like- 
wise III.  and  IV.,  and  likewise  V.  and  VI. ;  and  the  necessary 
formulas  are  (1)  Art.  36  (4)-(7)  or  (8)-(ll)  Art.  37,  and  (1),  (2), 
or  (3),  (4)  Art.  38.  Cases  V.  and  VI.  must  be  examined  as  to 
ambiguity;  and  accordingly,  they  give  more  trouble  than  the 
others.  Unless  the  triangle  satisfies  the  conditions  specified  in  Arts 
15,  17,  24  V.,  its  solution  is  impossible. 

Checks.  The  results  obtained  should  always  be  checked.  Delam- 
bre's  Analogies  and  formulas  which  have  not  been  used  in  the 
course  of  the  solution,  may  be  used  as  check  formulas. 

N.B.  Before  doing  any  of  the  numerical  work  the  student  should  try  to 
get  a  clear  idea  of  the  figure  of  the  triangle  upon  a  sphere.  This  geometrical 

54 


41-42.]  NUMERICAL   SOLUTIONS.  55 

conception  will  enable  him  to  make  a  reasonable  estimate  of  what  the  results 
will  be;  this  estimate  will  help  him  to  detect  wild  results  that  may  be 
obtained  by  making  numerical  errors.  For  example,  in  ABC  let  a  =  110°, 
6  =  114°,  C=  10°  ;  and  suppose  that  the  result  c= 76°  presents  itself.  A  person 
who  has  drawn  a  figure  of  the  triangle  on  a  sphere,  or  one  who  has  geometri- 
cal imagination  sufficient  to  give  him  an  idea  of  the  look  of  the  given  triangle, 
will  at  once  see  that  the  result,  c  =  76°,  must  be  wrong.  In  working  spherical 
triangles  it  is  much  better  not  to  proceed  blindly. 

42.    Case  I.     Given  the  three  sides. 

EXAMPLES. 
1.   In  ABC,  given  a  =  47°  30',  b  =  55°  40',  c  =  60°  10'.     Find  A,  B,  C. 


Jttnmdat:         ^nr^1™^^^ 


sin  t 
tan  r  tan  r 


tan     C= 


,  ,  . 

sin(s-a)  sin  (a  -6)  sin  (s  -  c) 

Check  :  Law  of  Sines,  or  Napier's  Analogies,  or  Delambre's  Analogies. 
Logarithmic  formulas  : 

log  tan2  r  -  log  sin  (s  -  a}  +  log  sin  (s  -  6)  +  log  sin  (s  -  c)  -  log  sin  s,  etc. 
Check  :  log  sin  a  —  log  sin  A  =  log  sin  b  —  log  sin  B  =  log  sin  c  —  log  sin  C. 

a  =    47°  30'                  log  sin  s  =    9.99539  -  10  .-.  £  A  =  28°  16'  2" 

b=    55°  40'  log  sin  (s  -  a)  =^74943^10^  %B  =  34°  33'  41.5" 

c=    60°  1W  log  sin  (s-  6)  =9.64184  -10  I  I  C  =  39°  29'  12" 

•.  2  s  =  163°  20'  log  sin  (s  -  c)  ^ko-SG^OSj-JB/  .'.  A  =  56°  32'  4" 

81°  40'            /.  log  tan2  r  =  18.95996  -  20  B  =  69°    T  23" 

34°  10'             /.  log  tan  r  =    9.47998  -  10  C  =  78°  58'  24" 

26°  /.  log  tan  |  A  =    9.73055  -  10 
21°  30'  log  tan  \  B  =    9.83814  -  10 

logtan£C  =    9.91590-10 

Check  :  log  sin  a  =  9.86763          log  sin  b  =  9.91686         log  sin  c  =  9.93826 

log  sin  A  =  Q.  92  128         log  sin  J5  =  9.97051        log  sin  C  =  9.99191 

9.94635  9.94635  9.94635 

NOTE  1.  Directions  for  the  numerical  work:  Fill  in  the  first  column; 
turn  up  the  first  four  logarithms  in  the  second  column  (since  these  logarithms 
are  required  by  the  formulas)  ;  compute  the  last  five  logarithms  in  the  second 
column  according  to  the  formulas  (these  computations  may  be  made  on 
another  paper,  if  necessary)  ;  find  the  first  three  angles  of  the  third  column 
by  the  tables  ;  thence  compute  A,  B,  C. 

NOTE  2.  If  only  one  angle  is  required,  say  A,  it  can  be  found  by  one  of 
formulas  (4)  Art.  37  ;  preferably,  the  second.  Angle  A  can  also  be  founcl 
(without  logarithms)  by  formula  (1)  Art.  37, 


56  SPHERICAL    TRIGONOMETRY.  [Cn.  IV. 

X  2-  Solve  ABC,  given  that  a  =  43°  30',  b  =  72°  24',  c  =  87°  50'. 

3.  Solve  ABC,  given  that  a  =  110°  40',  b  =  45C  10',  c  =  73°  SO'. 

4.  Solve  ABC,  given  that  a  =  120°  50',  6  =  98°  40',  c  =  74°  60'. 

5.  Solve  PQR,  given  that  p  =  67°  40',  q  =  47°  20',  r  =  83°  50'. 

43.  Case  II.     Given  the  three  angles. 

Either:  Solve  the  polar  triangle  by  the  method  used  in  Case  L, 
and  therefrom  obtain  the  parts  of  the  original  triangle. 
Or:  Solve  by  means  of  formulas  (8)-(ll)  Art.  37. 

EXAMPLES. 

Solve  ABC,  and  check  the  results. 
L^\.    Given  A  =  74°  40',  B  =  67°  30',  C  =  49°  50  . 

2.  Given  A  =  112°  30',  B  =  83°  40',  C  =  70°  10'. 

3.  Given  A  =  130°,  B  =  98°,  C  =  64°. 

4.  Given  P  =  33°  40',  $  =  26°  10',  R  =  20°  30'.     Find  p,  q,  r. 

NOTE.     The  results  may  also  be  checked  by  solving  the  examples  by  both 
the  methods  above. 

44.  Case  III.     Given  two  sides  and  their  included  angle. 

EXAMPLES. 
1.    In  ABC,  a  =  64°  24',  b  =  42°  30',   C  =  58°  40'  ;  find  A,  B,  c. 


Formulas  :  tan  $(A  +  B)  =          -n      Cot  \  C  ; 

cos    (a  +  6) 


sin  i  O  +  6) 

sin  c  =  5*™  Sin  a 
sin  ^4 

Checks  :  Law  of  Sines,  etc. 

C  =  58°  40'  log  cot  £C=0.25031         .-.  log  tan  ^(A  +  B)=  0.46743 

a=  64°  24'         log  sin  i  (a  +  6)  =9.90490  -10      log  tan  \(A-B)  =9.62405 
b  =  42°  30'        log  cos  £(«  +  &)  =9.77490-  10  •'•  K-4+-B)  =  71°  10'  41" 

...  a-}-  6  =106°  54'        logsin  £(«-  6)  =9.27864-10..  l(A-B)=22°W  12" 

a-&=21°54'        log  cos  $(a-6)=9.  99202  -10  /.  ^1=93°  59'  53" 

|C=  29°  20'  logsina=9,95513-10  £=48°  21'  29" 

J(a+6)=  53°27'  logsin  0=9.93154-10  .-.  log  sin  .4=9.  99894  -1( 

J(a-6)=  10°  57  /.  log  sine  =9.  88773  -1( 

/.  c=60°33'6'' 


43-46.]  THE  AMBIGUOUS  CASE.  57 

NOTE  1.     Since  C<A,  then  c<a  ;  and  hence,  the  acute  value  of  c  is  taken. 

NOTE  2.  Directions  for  the  numerical  work:  Fill  in  the  first  column; 
then  turn  up  all  the  logarithms  for  the  second  column,  these  logarithms  being 
required  by  the  formulas  ;  then  compute  the  first  two  logarithms  in  the  third 
column,  according  to  the  formulas  ;  thence  find  the  corresponding  angles,  and 
calculate  A  and  B ;  turn  up  log  sin  A ;  compute  log  sin  c  according  to  the 
formula ;  then  find  c  in  the  Tables. 

NOTE  3.  In  using  formulas  involving  the  difference  of  two  sides  or  two 
angles,  place  the  larger  side  or  angle  first. 

L   2.  Solve  ABC,  given  a  =  93° 20',  &=    56° 30',   C  =  74° 40'. 

3.  Solve  ABC,  given  b  =  76°  30',  c  =    47°  20',  ^  =  92°  30'. 

4.  Solve  ABC,  given  c  =  40°  20',  a  =  100°  30',  B  =  46°  40'. 

5.  Solve  PQE,  given  #  =  76°  30',  r  =  110°20',  P=46°50'. 

45.   Case  IV.     Given  one  side  and  its  two  adjacent  angles. 

Either :   Solve  the  polar  triangle  by  the  method  used  in  Case  III.  ;  and 
therefrom  obtain  the  parts  of  the  original  triangle. 
Or  :   Solve  by  using  formulas  (1),  (2),  Art.  38. 


EXAMPLES. 
1.    Solve  ABC,  given  A  =  67°  30',  B  =  45°  50',   c  =  74°  20f . 

2.  Solve  ABC,  given  I?  =  98°  30',   C  =  61°  20',  a  =  60°  40'. 

3.  Solve  ABC,  given  C  =  110°,       A  =  94°,         b  =  44°. 

4.  Solve  PQE,  given  E  =  70°  20',  Q  =  43°  50',  p  =  50°  46'. 

46.   Case  V.     Given  two  sides  and  the  angle  opposite  one  of  them. 

This  is  an  ambiguous  case,*  since  (Art.  24,  V.)  there  may  be  two 
solutions.  It  may  be  well  to  examine  this  case  (1)  geometrically, 
that  is,  by  an  inspection  of  the  figure ;  (2)  analytically,  that  is, 
by  an  inspection  of  the  formulas  involved  in  its  solution. 

(1)  Geometrically.  In  Art.  24,  V.  (Figs.  21,  25)  it  has  been 
seen  that,  when  two  sides  and  an  angle  opposite  one  of  them 
(say,  a,  b,  A)  of  a  triangle  ABC  are  given,  there  are  two  triangles 
possible  if  either  of  the  following  sets  of  conditions  holds,  viz. : 

A  <  90°,  a  >p,  a  <  b,  and  a  <  180°  -  b ;  (a) 

A  >  90°,  a  <  p,  a>b,  and  a  >  180°  -  b.  (b) 

*  For  a  detailed  discussion  of  the  ambiguous  case,  see  Todhunter,  Spher- 
ical Trigonometry,  pp.  53-58;  M'Clelland  and  Preston,  Spherical  Trigo- 
nometry, pp.  137-143. 


58  SPHERICAL   TRIGONOMETRY.  [Cn.  IV. 

In  order  that  the  triangle  be  possible,  it  is  apparent  that:  either 
CB=CP-,  or,  in  Fig.  21,  CB>CP,  i.e.  sin  CB  >  sin  CP,  i.e. 
sin  a  >  sin  AC  sin  CAP, 

i.e.  sin  a  >  sin  bsinA-,   . 

and,  in  Fig.  25.  CZ5  <  CP,  and  CLB  >  CP1,  i.e.  sin  a>CP',  i.e. 
sin  a  >  sin  AC  sin  CAP', 

i.e.  sin  a  >  sin  b  sin  (180°  —  CAP),  i.e.  sin  a  >  sm  b  sin  A. 

Art.  24  also  shows  that,  when  the  triangle  is  possible,  there  is 
one  solution  if  either  of  the  following  sets  of  conditions  holds,  viz.  : 

^4<90°,  a>p,  a  between  b  and  180°  -6;  (c) 

^>90°,  a<p,  a  between  b  and  180°  -  b.  (d) 

If  CB  =  CP,  i.e.  if  a  =  p,  then  there  is  one  solution. 
Art.  24  also  shows  that  the  triangle  is  impossible  if  either  one 
of  the  following  sets  of  conditions  holds,  viz.  : 

A  <  90°,  a  greater  than  both  b  and  180°  -  b  j  (e) 

A  >  90°,  a  less  than  both  b  and  180°  -  b. 

Since  the  greater  angle  is  opposite  the  greater  side,  B  must  be 
such  that  A  —  B  shall  have  the  same  sign  as  a  —  b. 

(2)  Analytically.  The  formulas  used  in  solving  this  case  are  as 
follows  : 


sma 


7|  cot  iC=!^|g±|  tan  l(A  -  B),     [or,  (4)  Art.  38]     (2) 


tan*(a~6>-     l>.  (2)  Art  38]     (3) 

Since  B  is  determined  from  its  sine,  it  may  be  in  either  the  first 
or  the  second  quadrant.  If  sin  a  —  sin  b  sin  A,  then  B  =  90°.  If 
sin  a  <  sin  b  sin  A,  then  sin  B  >  1,  and  B  has  an  impossible  value, 
and,  accordingly,  the  triangle  is  impossible.  [Compare  above.  J 
Equation  (2)  shows  that  A  —  B  and  a  —  b  have  the  same  sign. 


46.]  THE  AMBIGUOUS  CASE.  59 

Hence,  from  the  analytical  inspection  comes  the  following  rule: 

If  sin  a  <  sin  b  sin  A9  there  is  no  solution  ;  if  sin  a  =  sin  b  sin  A  , 
there  is  one  solution  ;  if  sin  a  >  sin  b  sin  JL,  cmcZ  if  both  values  of  B 
obtained  from  (1)  6e  swc/i  that 

A  —  B  and  a  —  b  have  like  signs, 

there  are  two  solutions;  if  only  one  o/  the  values  of  B  satisfies  this 
condition,  there  is  only  one  solution;  if  neither  of  the  values  of  B 
satisfies  this  condition,  the  solution  is  impossible. 

From  the  geometrical  inspection  comes  the  following  rule  : 

If  sin  a  <  sin  b  sin  A,  there  is  no  solution  ;  if  sin  a  -  sin  b  sin  A, 
there  is  one  solution  ;  if  sin  a  >  sin  b  sin  A9  then  : 

When  A  is  less  than  90°: 

there  are  two  solutions  if  a  is  less  than  both  b  and  180°  —  b; 
there  is  one  solution  if  a  lies  between  b  and  180°  —  b  ; 
there  is  no  solution  if  a  is  greater  than  both  b  and  180°  —  b. 

When  A  is  greater  than  90°: 

there  are  two  solutions  if  a  is  greater  than  both  b  and  180°  —  b; 
there  is  one  solution  if  a  lies  between  b  and  180°  —  b; 
there  is  no  solution  if  a  is  less  than  both  b  and  180°  —  b. 

NOTE  1.  The  second  rule  has  one  advantage  over  the  first,  in  that  it 
enables  one  to  say,  merely  on  calculating  sin  B,  but  without  finding  J5, 
whether  the  triangle  is  ambiguous  or  not. 

NOTE  2.  The  property  observed  in  Art.  36,  Note  2,  is  frequently  used  in 
investigating  the  ambiguous  case. 

EXAMPLES. 
1.    In  ABC,  a  =  43°  20',  6  =  48°  30',  A  =  58°  40'  ;   find  B,  C,  c. 


Formulas:  sin  B  =. 

sin  a 


C0t  *  °  =  ta"  *  (B  -  A)  •  [Art  38' 


tan  *  "  =  ri-       ta"  *  (6  -  a)'  [Art'  38' 

Checks:  Formulas  (2),  (4),  Art.  38  ;  or,  formulas,  Art.  37  ;  or,  Delambre's 
Analogies. 


60  SPHERICAL   TRIGONOMETRY.  [Cn.  IV. 

A  =  58°  40'  log  sin  A  =  9.93154  -  10  .-.  B  +  A  =  127°  27' 

a  =  43°  20'  log  sin  a  =  9.83648  -  10  B  -  A  =    10°   7' 

6  =  48°  30'  log  sin  b  =  9.87446  -  10  %(B  +  A)  =    63°  43'  30" 

.-.  b  +  a  =  91°  50'  .-.  log  sin  B  =  9.96952  -  10  %(B-A)=      5°   3'  30" 

b-a=    5°  10'  .-.  B  =    68° 47'  .-.  B'  +  A  =  169° 53' 

£(&  +  «)  =45°  55'  .B'  =  lll°13'  B'-A=    52°  33' 

£(&  -  a)  =    2°  35'  [According  to  the  test  for  $(B'  +  A)  =    84°  56'  30" 

ambiguity.]  $(B'  -  A)  =   26°  16'  30" 

In  ABC.          (See  Fig.  21,  Art.  24.)          In  AB'C. 

log  sin  £(&  +  a)  =  9. 85632  -10  r 

log  sin  £(&  -  a)  =  8.65391  -  10  j  As  in  ABC. 
log  tan  J(6  -  a)  =  8.65435  -  10 

log  sin  J(B  +  A)  =  9.95264  -  10  log  sin  %(B'  +  A)  =  9.99830  -  10 

log  sin  J(B  -  -4)  =  8.94532  -  10  log  sin  $(.B'  -  A)  =  9.64609  -  10 

log  tan  %(B  -  A)=  8.94702  -  10  log  tan  J(J3'  -  ^1)  ='  9.69345  -  10 

/.  log  cot  J  C  =  0. 14943  .-.  log  cot  \  C  =  0.89586 

log  tan  £  c  =  9.66167  -  10  log  tan  J  c  =  9.00656  -  10 

.-.  £C=35°19'55".4,  £c=24°38'53".  I   .-.  %C=  7°  14' 36",    Jc=  5° 47' 49". 

.-.     (7=  70° 39' 51",        c=49°17'46".  I   .-.     O  =  14°29' 12",       c=ll°35'38". 

Hence,  the  solutions  are  : 

ABC=    68°  47',    ACS  =  70°  39'  51",    AB  =  49°  17'  46"  ; 
AB'  (7=111°  13',  ACS'  =  14°  29'  12",  AB1  =  11°  35'  38". 

NOTE  3.  Directions  for  the  numerical  work :  Fill  in  the  first  of  the  three 
columns ;  turn  up  the  first  three  logarithms  in  the  second  column,  these 
being  required  by  the  first  formula ;  compute  log  sin  B  according  to  the  first 
formula  ;  find  B  in  the  tables  ;  decide  the  question  of  ambiguity ;  fill  in  the 
third  column  (only  four  lines  when  the  triangle  is  not  ambiguous).  Turn  up 
the  first  six  logarithms  in  the  first  of  the  next  two  columns ;  compute  the 
next  two  logarithms  according  to  the  formulas  ;  find  the  corresponding  values 

\     in  the  Tables  ;  thence  compute  C  and  c.     If  the  case  is  ambiguous,  do  the 

V  same  work  for  the  second  triangle. 

2.  Solve  ABC  when  a  =  56°  40',  b  =    30°  50',  A  =  103°  40'. 

3.  Solve  ABC  when  a  =  30°  20',  b  =    46°  30',  A  =    36°  40'. 

4.  Solve  ABC  when  c  =  74°  20',  a  =  119°  40',   C  =    88°  30'. 

5.  Solve  ABC  when  b  =  30°  10',  c  =    44°  30',  B  =    86°  50'. 

6.  Solve  PQR  when  q  =  42°  30',  r=   46°  50',   Q=    56°  30'. 


47-48.]  SUBSIDIARY  ANGLES.  61 

47.  Case  VI.    Given  two  angles  and  the  side  opposite  one  of  them. 

This  is  also  an  ambiguous  case. 

Either :  Solve  the  polar  triangle  by  the  method  used  in  Case  V. ; 
and  therefrom  obtain  the  parts  of  the  original  triangle. 

Or :  Solve  by  using  formula  (1)  Art.  36,  and  Napier's  Analogies. 

The  first  rule  (Art.  46)  for  determining  ambiguity  suits  the 
case,  if  a,  6,  be  substituted  for  A,  B,  therein.  On  making  use  of 
the  polar  triangle,  it  is  found  that  the  second  rule  can  be  adapted 
by  substituting  a,  A,  B,  for  A,  a,  b,  respectively. 

EXAMPLES. 

I/I.  Solve  ABC  when  A  =  108°  40',  B  =  134°  20',  a  =  145°  36'. 

2.  Solve  ABC  when  B  =  36°  20',     C  =  46°  30',  b  =  42°  12'. 

3.  Solve  ABC  when  C  =  62°  10',     A  =  23°  46',  c  =  33°  50'. 

4.  Solve  8TV  when  T  =  102°  50',   F=  81°  20',  t  =  124°  30'. 

48.  Subsidiary  angles.      Formulas  can  sometimes  be  adapted  for  loga- 
rithmic computation  and  the  triangle  solved,  by  the  use  of  subsidiary  angles. 
For  example,  in  ABC  let  a,  c,  B  be  known,  and  b  required.     (See  Fig.  35, 
Art.  33.) 

cos  b  =  cos  a  cos  c  +  sin  a  sin  c  cos  B  (Art.  36,  B) 

—  cos  c  (cos  a  -f  sin  a  tan  c  cos  J5). 
On  putting  tan  c  cos  .B  =  tan  0,  this  becomes 

cos  b  =  cos  c  (cos  a  +  sin  a  tan  0) 

__  cos  c  (cos  a  cos  0  +  sin  a  sin  0) 
cos  0 

_  cos  c  cos  (q  —  0) 
cos  0 

On  referring  to  Fig.  35  it  is  seen  that  BD  =  0,  that  DC  =  a  —  0,  and 

cos  .4Z)  = ;  so  that  solving  as  above  is  equivalent  to  solving  the  triangle 

cos  0 
by  dividing  it  into  right-angled  triangles. 

N.B.    Questions  and  exercises  on  Chapter  IV.  will  be  found  on  page  105. 


CHAPTER   V. 

CIRCLES  CONNECTED  WITH  SPHERICAL  TRIANGLES. 

49.  The  circumscribing  circle.  The  circle  passing  through  the 
vertices  of  a  spherical  triangle  is  called  the  circumscribing  circle, 
or  drcum-circle,  of  the  triangle.  This  circle  can  be  constructed 
in  somewhat  the  same  manner  as  the  circumscribing  circle  of  a 
plane  triangle. 

Let  ABC  (Fig.  41)  be  a  spherical  triangle,  and  let  R  denote 
the  radius  (i.e.  the  polar  distance,  Art.  6) 
of  its  circumscribing  circle.  Bisect  the 
arcs  BC,  CA,  in  L,  M,  respectively;  and 
at  Z,  M,  draw  arcs  at  right  angles  to  BC, 
CA,  respectively.  The  point  0,  at  which 
these  arcs  meet,  is  the  pole  of  the  circum- 
scribing circle. 

For,  draw  OA,  OB,   OC,  arcs  of  great 
circles.     In  the  triangles  OLB  and  OLC, 

BL  =  LC,  LO  is  common,  and  the  angles  at  L  are  right  angles. 
Hence,  OB  =  OC.  In  a  similar  way  it  can  be  shown  that 
OC  =  OA.  Hence  0  is  the  pole  of  the  circumscribing  circle. 

Join  0  and  N,  the  middle  point  of  AB ;  then  it  is  easily  shown 
that  ON  is  at  right  angles  to  AB. 

In  ABC,  A  +  B+C=2S. 

Now  (since  OA  =  OB  =  OC), 

OAB=OBA,  OBC=OCB,  OCA=  OAC. 
Hence,  OAB+ OBC+ OAC=S. 

...   OBC=  S-(OAB  +  OAC)  =  S-  A. 
In  the  right-angled  triangle  OBL, 

^°B-^L-         C^t.  26,  E,,  (3)] 
62 


49-50.] 


THE  INSCRIBED   CIRCLE. 


63 


On  substituting  in  (1)  the  value  of  tan  J  a  in  relation  (8)  Art. 
37,  equation  (1)  becomes 


tan 


-V 


—cos 


cos  (#  -  A)  cos  (£  -  B)  cos  (S—C) 


NOTE  1.  Compare  (1)  with  the  corresponding  case  in  plane  triangles 
(Plane  Trig.,  Art.  68).  (In  plane  triangles,  £=90°,  and,  hence, 
COB  (#--4)  =  sin  4.) 

NOTE  2.     On  putting    N=  V-  cos  /SYcos(#  -  A)cos(S  -  B)cos(S  -  (7), 


50.  The  inscribed  circle.  The  circle  which  touches  each  of  the 
sides  of  a  spherical  triangle  is  called  the  inscribed  circle,  or  in- 
circle,  of  the  triangle.  This  circle  can  be  constructed  in  some- 
what the  same  manner  as  the  inscribed  circle  of  a  plane  triangle. 

Let  ABC  be  a  spherical  triangle,  and  let 
r  denote  the  radius  (f..e.  the  polar  distance) 
of  its  inscribed  circle.  Bisect  angles  A,  B, 
by  arcs  of  great  circles,  and  let  these  arcs 
meet  at  0.  Draw  OL,  OM,  ON,  at  right 
angles  to  BC,  CA,  AB,  respectively. 

In  the  triangles  0AM  and  OAN,  the 
angles  at  A  are  equal,  the  angles  at  N  and 
M  are  right  angles,  and  the  side  OA  is 
common.  Hence  these  triangles  are  symmetrical,  and  OM  '=  ON. 
Similarly  it  can  be  shown  that  0^7"=  OL.  Hence  0  is  the  pole 
of  the  circle  inscribed  in  ABC. 

Since  the  triangles  0AM  and  OAN  are  equal,  AM  =  AN. 
Similarly,  BN=BL,  and  CL  =  CM. 

Now  AB  -}-  BC  +  CA  =  2  s  • 

hence  AN  +  BL  +  CL  =  s. 

.-.  AN=  s  -  (BL  +-LC)  =  s  -  a. 


64                             SPHERICAL   TRIGONOMETRY.  [Cn.  V. 
In  the  right-angled  triangle  AON, 

tan  ON  =  tan  OAN  sin  AN.  [Art.  26,  (4)] 

.'.  tan  r  =  tan  |  A  sin  (s  -  a) .  (1) 

Similarly,  tan  r  —  tan  £  B  sin  (s  —  6)  ;   tan  r  =  tan  1  (7  sin  (s  —  c). 

On  substituting  in  (1)  the  value  of   tsm^A  in  (4)  Art.  37, 
equation  (1)  becomes 


tan  r 


=  A/  sin  O  •— 
* 


sin  * 

On  putting     n  =  Vsin  s  sin  (s  -  a)  sin  («  -  ft)  sin  («  -  c)  ,  (3) 

tanr  =  -^-.  (4) 

sin  s 

NOTE  1.  Compare  (1)  with  Plane  Trigonometry,  Art.  69,  Note  ;  (2)  with 
Art.  69,  (3);  n  with  S,  Art.  66,  (3);  (4)  with  (3)  Art.  69. 

51.  Escribed  circles.  A  circle  which  touches  a  side  of  a  spher- 
ical triangle,  and  the  other  two  sides  produced  (that  is,  which 
is  inscribed  in  a  co-lunar  triangle),  is  an  escribed  circle,  or  an  ex- 
circle,  of  the  triangle.  There  are  three  ex-circles,  one  correspond- 

ing to  each  side  of  the  triangle. 

Let  ABC  be  a  spherical  tri- 
angle ;  and  let  the  radii  of  the 
escribed  circles,  touching  a,  b,  c, 
respectively,  be  denoted  by  ra,  rb, 
rc,  respectively.  Complete  the 
lune  whose  angle  is  A.  The 

escribed  circle  which  touches  a   is   the  inscribed   circle  of  the 
co-lunar  triangle  ABC.     Hence  [Art.  50,  (1)], 


tan  ra  =  tan  £  A'  sin  }  [(a  +  180°  -  b  +  180°  -  c)  -  2  a]  ; 
i.e.  tan  ra  =  tan  |  A  sin  s.  (1) 

Similarly,  tan  r6  =  tan  ±  B  sin  s ;  tan  rc  =  tan  £  C  sin  s. 


51.]  ESCRIBED   CIRCLES.  65 

On  substituting  for  tanjyl  its  value  in  (4)  Art.  37,  equation 
(1)  becomes 

tan  r  = 


sin  (s  —  a) 
tann-  =  [^t.  50,  (3)]  (3) 


Similarly,     tan  rb  =  — — -^ ;   tan  rc  =  — 


sin  (s  —  6)  sin  (s  —  c) 

NOTE.  Compare  (3)  with  the  corresponding  result  in  Plane  Trigonome- 
try, Art.  70. 

Some  other  relations  between  the  sides  and  angles  of  a  spherical 
triangle  and  the  radii  of  the  circles  connected  with  it}  are  indicated 
in  the  exercises  at  the  end  of  the  book. 

Ex.  Find  the  radii  of  the  circumscribing,  inscribed,  and  escribed  circles  of 
some  of  the  triangles  in  Chapters  II.,  IV. 

N.B.     For  questions  and  exercises  on  Chapter  F.,  see  page  107. 


CHAPTER   VI. 


AREAS  AND   VOLUMES   CONNECTED  WITH   SPHERES. 

52.   Preliminary  propositions. 

a.  The  lateral  area  of  a  frustum  of  a  regular  pyramid  is  equal 
to  the  product  of  the  slant  height  of  the  frustum  and  half  the 
sum  of  the  perimeters  of  its  bases. 


Ps 


re,, 

LK-'-iO 


J  T> 

Fia.  45 


FlO. 


The  student  can  easily  prove  this  (Fig.  44).  It  should  be 
noted  that  the  half  sum  of  the  perimeters  of  the  bases  of  the 
frustum  is  equal  to  the  perimeter  of  the  section  which  is  parallel 
to  the  bases  and  midway  between  them. 

In  symbols:  If  p^  p.2,  P,  are  the  perimeters  of  the.  bases  and 
the  middle  section  of  the  frustum,  and  MN  is  its  slant  height, 
theii 

lateral  area  of  frustum  =  1  MN  (pl  +  p>l)  =  MN  -  P. 

6.  The  lateral  area  of  a  frustum  of  a  cone  of  revolution  is 
equal  to  the  product  of  the  slant  height  of  the  frustum  and  half 
the  sum  of  the  circumferences  of  its  bases. 

[Suggestion  for  proof:  If  the  number  of  the  lateral  faces  of  a 
frustum  of  a  regular  pyramid  be  indefinitely  increased  and  each 
face  be  indefinitely  decreased,  then  this  frustum  approaches  the 
frustum  of  a  cone  of  revolution  as  a  limit  (see  Fig.  46).  Accord- 
ingly, Proposition  (b)  follows  at  once  from  (a)].  It  should  be 


52-53.]  AREA    OF  A   SPHERE.  67 

noted  that  half  the  sum  of  the  circumferences  of  the  bases  of. 
the  frustum  is  equal  to  the  circumference  of  the  section  which 
is  parallel  to  the  bases  and  midway  between  them. 

In  symbols  :  If  Q,  C2,  C  (Fig.  45)  are  the  circumferences  of  the 
bases  and  the  middle  section  of  the  frustum,  and  MN  is  its  slant 
height,  then  lateral  area  of  frustum 


=  |  MN  (Ci  +  Ci)  =  MN-  C=2  TrLG  -  MN. 

NOTE:  The  lateral  surface  of  the  frustum  of  the  cone  (Fig.  45)  can  be 
generated  by  the  revolution  of  the  line  MN  about  the  line  AB  which  is  in 
the  same  plane  with  MN. 

53.  To  find  the  area  of  a  sphere.  The  surface  of  a  sphere  can 
be  generated  by  the  revolution  of  a  semicircle  about  its  diameter. 
For  example,  the  semicircle  ATKB  of  radius  R 
on  revolving  about  its  diameter  AB,  will  describe 
the  surface  of  a  sphere  of  radius  OA. 

Let  a  polygon  ALTGKB  be  inscribed  in  this 
semicircle.  At  M,  the  middle  point  of  one  of  the 
chords  LT,  draw  MO  at  right  angles  to  LT.  By  Q 
geometry,  MO  will  meet  AB  at  0,  the  middle 
point  of  AB.  Project  LT  on  AB,  the-  projection 
being  It  ;  draw  LQ  at  right  angles  to  Tt. 

By  Art.  52.  6,  the  area  generated  by  LT  in  its  revolution 
about  AB 

=  2  irMm.LT.  (1) 

Since  the  angles  of  the  triangle  LTQ  are  respectively  equal  to 
the  angles  of  OMm,  these  triangles  are  similar  ;  accordingly, 

LT:LQ=  OM:Mm. 
.-.  Mm-LT=LQ.  OM  =  It  .  OM. 


Hence,  from  (1),  area  generated  by  LT  =  2  vOM  •  It.  (2) 

In  words:  When  a  chord  of  a  semicircle  revolves  about  the 
diameter,  the  area  generated  is  equal  to  2  TT  times  the  product  of 
the  length  of  the  perpendicular  from  the  centre  to  the  chord, 
and  the  projection  of  the  chord  upon  the  diameter. 


68  SPHERICAL    TRIGONOMETRY.  [Cn.  VI. 

.*.  The  area  of  the  surface  generated  by  the  revolution  of  the 
polygon  ALTGKB 

=      2  TT  x  (perpendicular  on  AL  from  0)  X  Al 
-|-  2  TT  x  (perpendicular  on  LT  from  0)  x  It 
+  2  TT  x  (perpendicular  on  TG  from  0)  x  tg 
4-  2  TT  x  (perpendicular  on  GKfrom  0)  x  #& 
+  2  TT  x  (perpendicular  on  KB  from  0)  x  kB. 

If  the  number  of  sides  in  the  polygon  inscribed  in  the  semi- 
circle is  indefinitely  increased  and  each  side  is  indefinitely  de- 
creased, then  the  broken  line  ALTGKB  approaches  the  semicircle 
as  a  limit,  and  each  of  the  perpendiculars  drawn  from  0  to  the 
middle  points  of  the  chords  approaches  R  as  a  limit  ;  while 
the  sum  of  the  projections  of  the  chords  remains  equal  to  AB, 
the  diameter  of  the  circle.  Hence,  area  of  surface  generated  by 
revolution  of  semicircle  AGB  =  2  -*•  •  R  •  2  R  ; 

i.e.  area  of  surface  of  sphere  of  radius  R  =  4  wR2. 

In  words  :  The  area  of  the  surface  of  a  sphere  is  four  times  the 
area  of  a  great  circle  of  the  sphere. 

Definition.  A  zone  of  a  sphere  is  a  portion  of  the  surface  in- 
cluded between  two  parallel  planes,  or,  what  comes  to  the  same 
thing,  is  the  portion  of  the  surface  included  between  two  circles 
which  have  common  poles  ;  for  example,  the  surface  between  the 
parallels  of  30°  N.  latitude  and  50°  K  latitude. 

The  area  of  a  zone.  An  infinite  number  of  chords  can  be  in- 
scribed in  the  arc  LT  (Fig.  47).  By  reasoning  similar  to  that 
employed  above,  it  can  be  shown  that 


area  of  surface  generated  by  arc  LT=  2  irR  •  It. 

.-.  The  area  of  a  spherical  zone  is  equal  to  the  product  of  the  length 
of  a  great  circle  of  the  sphere  and  the  height  of  the  zone. 

It  follows  that  on  a  sphere  or  on  equal  spheres  the  areas  of 
zones  of  equal  heights  are  equal. 


54-55.]  A   SPHERICAL  DEGREE  DEFINED.  69 

EXAMPLES. 

1.  Find  the  area  of  a  sphere  of  radius  15  inches. 

2.  Find  the  surface  of  a  spherical  zone  of  height  2.5  inches  on  a  sphere  of 
diameter  50  inches. 

3.  Find  the  convex  surface  of  a  spherical  segment  of  height  4.5  inches  on 
a  sphere  of  diameter  7  feet.     [See  definition,  Art.  63.] 

4.  Suppose  that  the  earth  is  a  sphere  whose  radius  is  3960  miles  ;  find  the 
area  of  the  surface  included  between  the  North  Pole  and  the  parallel  of  80° 
N.  latitude  ;  between  the  parallels  of  49°  N.  and  50°  N.  ;  between  6°  N. 
and  5°  S. 

54.  Lunes.     Definition.     The  spherical  surface  bounded  by  two 
halves  of  great-circles  is  called  a  lune;  e.g.  the  surface  between 
two  meridians.     The  angle  of  the  lune  is  the  angle  between  the 
two    semicircles ;    thus    the   angle   of    the 

lune    between    the   meridians    70°  W.   and 
80°  W.  is  10°. 

Proposition.  On  the  same  circle  or  on 
equal  circles  the  areas  of  lunes  are  propor- 
tional to  their  angles.  This  can  be  proved 
by  a  method  similar  to  that  which  is  used 
in  proving  that  the  angles  at  the  centre  of 
a  circle  are  proportional  to  the  arcs  sub-  FIG.  48 

tended  by  them. 

55.  A  spherical  degree  defined.     From  the  proposition  in  Art.  54 
it  follows  that  the  area  of  a  lune  is  to  the  area  of  the  surface 
of  the  sphere  as  the  angle  of  the  lune  is  to  four  right  angles. 

That  is, 

area  of  lune  of  angle  A°  :  area  of  sphere  =  A°  :  360°. 

Hence,  area  of  lune  of  angle  1°  =  area  of  sPhere. 

360 

Let  a  great  circle  be  drawn  about  one  of  the  vertices  of  a  lune 
of  angle  1°  as  a  pole.  The  lune  is  then  divided  into  two  equal  bi- 
rectangular  triangles ;  accordingly,  each  triangle  contains  (7J^)th 
of  the  surface  of  the  sphere,  or  (^^th  of  the  surface  of  the 
hemisphere.  The  surface  of  each  such  triangle  is  called  a  spherical 
degree. 


70  SPHEEICAL    TRIGONOMETRY.  [Cn.  VI. 

For  example,  the  part  of  the  surface  of  a  globe  bounded  by  the 
meridians  43°  W.  and  63°  W.  longitude  and  the  equator,  contains 
20  spherical  degrees ;  the  lune  bounded  by  these  meridians  con- 
tains 40  spherical  degrees. 

A  lune  of  angle  A°  contains  2  A  spherical  degrees. 

The  passage  from  spherical  degrees  of  surface  to  the  ordinary 
measure  (of  the  area)  of  the  surface  is  easily  effected  when  the 
radius  of  the  sphere  is  given. 

A  spherical  degree  =  (^^)tli  part  of  the  surface  of  a  sphere; 
hence,  on  a  sphere  of  radius  r, 

a  spherical  degree  contains  ^TT'      i-e-  -  •—  square  units  of  area. 

Thus, 

area  of  a  lune  of  angle  20°  on  a  sphere  of  radius  r  =  —     —  =  f  Trr2. 

180 

EXAMPLES. 

1.  Find  the  area  of  a  lune  of  angle  10°  on  a  sphere  of  radius  2  feet. 

2.  Find  the  area  of  a  lune  of  angle  37°  30'  on  a  sphere  of  radius  7  feet. 

3.  Find  the  area  between  the  meridians  77°  W.  and  83°  20'  W. ;  and  the 
area  between  the  meridians  174°  20'  W.  and  158°  35'  E.     (Radius  of  earth 
=  3960  miles.)     [Express  areas  in  spherical  degrees  and  in  square  miles.] 

56.  Spherical  excess  of  a  triangle.  The  sum  of  the  angles  of  a 
plane  triangle  is  always  equal  to  180° ;  the  sum  of  the  angles  of 
a  spherical  triangle  is  always  greater  than  180°  (Art.  17).  The 
difference  between  the  latter  sum  and  180°  is  called  the  spherical 
excess  of  the  triangle.  (This  excess  is  due  to  the  fact  that  the 
triangle  is  spherical  and  not  plane;  hence  the  excess  is  called 
spherical.)  For  example,  in  the  triangle  bounded  by  the  meridi- 
ans 47°  W.  and  48°  W.  longitude  and  the  equator,  the  sum  of  the 
angles  is  181° ;  and,  accordingly,  the  spherical  excess  is  1°.  In 
the  triangle  bounded  by  the  meridians  43°  W.  and  63°  W.  and 
the  equator  the  sum  of  the  angles  is  200°,  and  the  spherical  ex- 
cess is  20° ;  in  the  spherical  triangle  having  angles  50°,  65°,  125°, 
the  spherical  excess  is  (50°  +  65°  +  125°  -  180°),  i.e.  60°. 


56-57.] 


AREA    OF  A    SPHERICAL    TRIANGLE. 


71 


If  E  denote  the  number  of  degrees  in  the  spherical  excess,  and 
Er  denote  the  number  of  radians  therein,  then 


in  a  triangle  ABC,    E°  =  A°  +  B°  +  C°  -  180° ; 
and  [Plane  Trigonometry,  Art.  73,  (7)], 


180 


(1) 


(2) 


Ex.  Find  the  spherical  excess  (in  degrees  and  in  radians')  of  the  tri- 
angles described  in  Art.  42,  Exs.  1,  2,  3  ;  Art.  43,  Exs.  1,  2  ;  Art.  44,  Exs. 
1,  2,  3  ;  Art.  45,  Exs.  1,  2;  Art.  46,  Exs.  1,  2,  3  ;  Art.  47,  Exs.  1,  2. 

57.   The  area  of  a  spherical  triangle. 

Proposition :  The  number  of  spherical  degrees  (of  surface)  in  a 
spherical  triangle  is  equal  to  the  number  of  (angular)  degrees  in 
its  spherical  excess* 

Let  ABC  be  a  spherical  triangle  whose 
spherical  excess  is  E° ;  then  area  ABC 
is  equal  to  E  spherical  degrees.  Com- 
plete the  great  circle  BCB'C',  and  pro-  / 
duce  the  arcs  BA,  CA  to  meet  this  circle 
in  B',  C',  respectively.  Complete  the 
great  circles  BAB'B  and  AC  AC'.  The 
triangle  AB'C'  is  equal  to  the  triangle 
ABC.  For, 


Fia.  49 


'A  =  180°-AC=CA, 
=  ISO°-B'C=CB. 
Hence,  in  area,  ABC  +  AB'C '  =  lime  ACA'BA; 
also  ABC  +  AB'C  =  lune  BCB'AB-, 

and  ABC  +  ABC'  =  lune  CBC'AC. 


*  This  proposition  is  sometimes  stated  thus :   The  area  of  a  triangle  is 
equal  to  its  spherical  excess  •  but  this  enunciation  is  rather  slipshod. 


72  SPHERICAL    TRIGONOMETRY.  [On.  VI. 

Hence,  on  addition, 
2  ABC  +  (ABC  +  AB'C'  +  AB'C  +  ^LBC") 

=  lime  ^4  -f-  lune  J5  +  lune  (7; 
2  ABC=  lune  ^4  +  lune  jB  +  lune  C  —  hemisphere. 
.-.  (by  Art.  55)  2  ABC  =  (2  A+2  B+2  C-360)  spherical  degrees. 
ABC  =  (A  +  5  +  (7  -  180)  spherical  degrees 
===  .JEJ  spherical  degrees. 

Since  (Art.  55)  a  spherical  degree  on  a  sphere  of  radius  r  con- 
tains T!"jy  ?rr2  square  units  of  area,  then,  on  this  sphere, 

area  ABC  =  A  +  B^"19°  -r»*  =  j^  -r*>  C1) 

=  J?rr2>  [Art.  56  (2)]     (2) 

in  which  1£  denotes  the  number  of  degrees,  and  Er  denotes  the 
number  of  radians  in  the  spherical  excess. 

Hence,  in  order  to  find  the  area  of  a  triangle,  find  the  angles, 
calculate  the  spherical  excess  in  degrees  or  radians,  and  use  one 
of  formulas  (1),  (2). 

NOTE.  It  should  be  observed  that  [from  Art.  14,  Art.  56  (1),  and  the 
proposition  above],  the  number  of  spherical  degrees  contained  in  the  area 
subtended  on  a  spherical  surface  by  a  solid  angle  at  the  centre  of  the  sphere, 
remains  the  same,  however  the  radius  may  vary.  On  the  other  hand,  by  (1) 
and  (2),  the  number  of  square  units  in  the  subtended  area  varies  as  the 
square  of  the  radius. 

*  This  expression  for  the  area  of  a  spherical  triangle  was  first  given  in 
1629  by  Albert  Girard  (1590-1634)  (see  Plane  Trigonometry,  pp.  22,  167); 
and  it  is  often  called  GirarcTs  Theorem.  The  method  of  proof  used  above 
was  invented  by  John  Wallis  (1616-1703)  professor  of  geometry  at  Oxford. 
(See  Wallis,  Work?,  Vol.  II.,  p.  875.) 

It  follows  from  (1)  that 

area  ABC  :  2  irR*  =  E°  :  360°. 


Hence,  the  above  proposition  may  be  expressed  thus  :  The  area  of  a 
spherical  triangle  is  to  the  surface  of  the  hemisphere  as  the  excess  of  its  three 
angles  above  two  right  angles  is  to  four  right  angles. 


58.]  SPHERICAL  EXCESS   OF  A    TRIANGLE.  73 

EXAMPLES. 
Find  the  areas  of  the  following  triangles  (see  examples,  Art.  56)  : 

1.  Those  described  in  Art.  42,  Exs.  1,  2,  3,  when  on  a  sphere  of  radius 
10  feet. 

2.  Those  described  in  Art.  43,  Exs.  1,  2,  when  on  a  sphere  of  radius 
25  inches. 

3.  Those  described  in  Art.  44,  Exs.  1,  2,  3,  when  on  a  sphere  of  radius 
30  yards. 

4.  Those  described  in  Art.  45,  Exs.  1,  2,  when  on  a  sphere  of  radius 
4  feet. 

5.  Those  described  in  Art.  46,  Exs.  1,  2,  3,  when  on  a  sphere  of  radius 
18  inches. 

6.  Those  described  in  Art.  47,  Exs.  1,  2,  when  on  a  sphere  of  radius 
3960  miles. 

58.  Formulas  for  the  spherical  excess  (£°)  of  a  triangle.  Since,  in 
a  spherical  triangle  ABC,  E°  =  A°  +  B°  -f  C°  -  180°,  and  since  there  are 
many  relations  between  the  sides  and  angles  of  a  triangle,  it  may  be  expected 
that  there  can  be  many  formulas  for  the  spherical  excess  ;  and,  accordingly, 
for  the  area  of  a  spherical  triangle.  [It  will  be  remembered  that  there 
are  several  formulas  for  the  area  of  a  plane  triangle  (Plane  Trigonometry, 
Art.  66).]  Following  are  some  of  the  most  important  of  these  (the  deduc- 
tion of  some  of  them  is  given  in  Note  JB)  : 

A.    The  spherical  excess  in  terms  of  the  three  sides. 
(a)  L^Huillier's  formula : 


tan  £ E°  -  Vtan  \ s  tan |(s  -  a)  tan £(s  -  6)  tan  |(s  -  c). 


(6)   CagnolVs  formula :  sin  \  E°  = 


n 


2  cos  \  a  cos  i  b  cos  \  c 


in  which  .  n  =  vsinssin  (s  —  a)sin(s  —  6)sin(s  —  c). 

(c)  De  Gun's  formula*:  cot  ^  =  l  +  COS(*  +  COB  b  +  COSC.f 


*  Simon  L'lluillier  (1750-1810),  a  Swiss  mathematician  and  philosopher ; 
Antoine  Cagnoli  (1743-1816),  an  Italian  astronomer;  L'abbe  Jean  Paul 
de  Gua  (1712-1786),  a  French  philosopher. 

t  For  the  deduction  of  this  formula  see  Chauvenet,  Trigonometry,  p.  230, 
and  Crawley,  Trigonometry,  p.  166. 


74  SPHERICAL   TRIGONOMETRY.  [On.  VI. 

B.    The  spherical  excess  in  terms  of  two  sides  and  their  included  angle. 

(d)  tan  i  E°  =  _J^Li^tan_^sin(7 

1  +  tan  I  a  tan  £  b  cos  O 

(e)  cot  i  ^°  -  cot  ^  a  cot  ^-f  cos  (7 

sin  O 

Ex.   By  these  formulas  find  the  spherical  excess  of  some  of  the  triangles 
referred  to  in  Ex.  1,  Art.  56. 

59.   a.  The  number  of  spherical  degrees  in  any  figure  on  a  sphere, 

whatever  may  be  its  boundary,  is  the  ratio  of  the  area  of  the 
figure  to  the  area  of  a  spherical  degree,  that  is,  to  (^J^)th  part  of 
the  area  of  the  hemisphere  (Art.  55).  Thus,  on  a  sphere  of  radius 
r,  if  A  denotes  the  area  of  the  figure,  and  E  the  number  of  spheri- 
cal degrees  therein,  then,  since  area  of  a  hemisphere  = 


[Compare  Art.  57  (1),  Art.  59  (2).] 

The  plane  angle  E°  may  be  called  the  spherical  excess  of  the 
figure.  For  example,  the  spherical  excess  of  a  lune  of  angle  A° 
is  2A°. 

b.  The  spherical  excess  of  a  (non-re-entrant)  spherical  polygon. 
On  drawing  diagonals  from  any  vertex  of  a  polygon  of  n  sides  to 
the  other  vertices,  it  will  be  seen  that  the  polygon  is  divided  into 
n  —  2  triangles.  The  sum  of  the  angles  of  all  these  triangles  is  the 
same  as  the  sum  of  the  angles  of  the  polygon.  Hence, 

spherical  excess  (E°)  of  polygon  ofn  sides 

=  sum  of  angles  -(n-2)  ISO0. 

If  the  radius  of  the  sphere  is  r,  then  (Art.  57) 

W 

area  of  the  polygon  =  -—  Trr2.  (2) 

J_Ow 

60.  Given  the  arsa  of  a  figure  :  to  find  its  spherical  excess.  More 
fully  :  To  find  the  spherical  excess  of  a  figure  on  a  sphere  when  the 
area  of  the  figure  is  given  in  square  units. 


59-60.]  SPHERICAL   EXCESS  OF  FIGURES.  75 

Let  r  denote  the  radius  of  the  sphere,  A  the  area  of  the  figure, 
E  the  number  of  degrees,  n  the  number  of  seconds,  and  Er  the 
number  of  radians,  in  its  spherical  excess.  Then,  by  (1)  Art.  59, 


.-.  n  =  3600  E  =  206265^,-  (2) 

Now  1°  =  -^-  radians  ; 

loU 

hence  E°  =  -~:E  radians 

180 

=  ^  radians.  [by  (1)] 

•'•  E'=f  (3) 

A  particular  application  of  (2)  can  be  made  to  the  following 
problem,  viz.  :  The  area  of  a  spherical  triangle  on  the  earth's  sur- 
face being  known,  to  derive  a  formula  for  computing  the  spherical 
excess. 

The  length  of  a  degree  on  the  earth's  surface  is  found  to  be 
365155  feet.  Accordingly, 

R  (the  radius  of  the  earth)  =  365155  x  18°  feet.  (4) 

Prom  (2),      log  n  =  log  A  +  log  206265  -  2  log  R.  (5) 

On  expressing  A  in  square  feet,  and  substituting  in  (5)  the 
value  of  R  in  (4),  there  is  obtained, 

log  n  =  log  A  -  9.3267737.  (6) 

Formula  (6)  is  called  Roy's  Rule,  as  it  was  used  by  General 
William  Roy  (1726-1790)  in  the  Trigonometrical  Survey  of  the 
British  Isles.*  The  area  of  the  spherical  triangle  can  be  approxi- 
mately determined  to  a  sufficient  degree  of  accuracy. 

*The  rule  should  probably  be  credited  to  Isaac  Dalby  (1744-1824),  who 
was  mathematical  assistant  to  General  Roy  from  1787  to  1790,  and  later 
became  professor  of  mathematics  at  the  Royal  Military  College.  [See  Phil. 
Trans.,  vol.  80  (1790).]  This  was  the  first  practical  application  of  Gerard's 
theorem  (Art.  57). 


76  SPHERICAL   TRIGONOMETRY.  [Cn.  VI. 

61.  The  measure  of  a  solid  angle.  A  plane  angle  can  be  measured 
by  any  circular  arc  which  it  subtends;  and  the  measure  can  be 
expressed  in  radians  and  in  degrees.  The  radian  (or  circular) 
measure  of  an  angle  is  the  number  of  times  any  circular  arc  sub- 
tended by  it  contains  the  radius  (Plane  Trig.,  Art.  73) ;  and  the 
number  of  degrees  in  the  angle  is  equal  to  the  number  of  degrees  in 
the  subtended  circular  arc.  Thus,  the  radian  measure  of  an  angle 
of  an  equiangular  triangle  is  ITT,  and  its  degree  measure  is  60. 

A  solid  angle  can  be  measured  in  a  somewhat  similar  manner, 
namely,  by  means  of  any  spherical  surface  which  it  subtends. 
What  may  be  called  the  spherical  measure  of  a  solid  angle  is  the 

number  of  times  any  spherical  sur- 
face subtended  by  it  contains  an 
area  equal  to  the  square  on  the  radius. 
For  example,  since  the  surface  of  a 
sphere  is  equal  to  4^,  the  sum  of 
all  the  solid  angles  about  any  point 
is  47T.  The  angle  at  the  corner  of 
a  cube  subtends  one-eighth  of  the 
FIQ.  50  surface  of  the  sphere ;  accordingly,  its 

spherical  measure  is  -^-  -f-  r2,  i.e.  ^  TT. 

A  solid  angle  may  also  be  measured  in  spherical  degrees,  a  term 
that  will  be  explained  presently.  What  may  be  called  the 
spherical  degree  measure  of  a  solid  angle  (or,  the  number  of 
spherical  degrees  in  the  angle)  is  a  number  equal  to  the  number 
of  spherical  degrees  of  area  in  any  spherical  surface  subtended 
by  the  angle.  An  angle  that  subtends  a  spherical  degree  of 
surface,  contains  what  may  be  called  a  solid  spherical  degree. 
For  example,  the  sum  of  all  the  solid  angles  about  any  point 
is  720  spherical  degrees  (of  angle);  the  angle  at  the  corner  of 
a  cube  contains  90  spherical  degrees  (of  angle).  Thus  the 
spherical  measure  of  the  angle  at  the  corner  of  a  cube  is  ^  TT,  and 
its  spherical  degree  measure  is  90.  On  comparing  these  defini- 
tions of  solid  angular  measures  with  Art.  55  and  equations  (3)  and 
(1)  Art.  60,  it  is  seen  that  these  measures  of  solid  angles  are  equal 
to  the  measures,  in  radians  and  degrees  respectively,  of  the  spherical 
excess  of  the  Jigures  subtended  on  any  sphere  by  the  angle,  when 
the  vertex  of  the  angle  is  at  the  centre  of  the  sphere. 


61.]  MEASUREMENT  OF  SOLID  ANGLES.  77 

NOTE  1.  The  term  degree.  In  geometry  and  trigonometry  the  word 
degree  is  used  in  connection  with  four  very  different  kinds  of  quantities ; 
namely,  circular  arcs,  plane  angles,  spherical  surfaces,  and  solid  angles. 

A  degree  of  arc,  or  an  arcual  degree,  is  (j^)th  part  of  any  circle  ; 

A  degree  of  angle,  or  an  angular  degree,  is  (^|^)th  part  of  four  right 
angles ; 

A  degree  of  surface  on  a  sphere,  or  a  spherical  degree  of  surface,  is  (7^)th 
part  of  the  surface  of  any  sphere  ; 

A  degree  of  solid  angle,  or  a  solid  spherical  degree,  is  (7^)th  part  of  the 
solid  angles  about  any  point. 

NOTE  2.  If  two  plane  angles  are  equal,  they  can  be  superposed,  the  one 
on  the  other.  On  the  other  hand,  just  as  two  figures  on  a  sphere  may  be 
equal  in  area  and  differ  in  every  other  respect,  so  two  solid  angles  can  be 
equal  in  measure  and  differ  in  every  other  respect. 

NOTE  3.  The  following  remarks  relating  to  the  measurement  of  solid 
angles  are  from  Hutton's  Course  in  Mathematics,  Vol.  II.,  p.  64  : 

"  Solid  angles :  If  about  the  angular  point  of  a  solid  angle  as  centre,  a 
sphere  be  described  to  radius  unity,  the  portion  of  its  surface  intercepted 
between  the  planes  which  contain  the  solid  angle  is  the  measure  of  the 
solid  angle.  (This  method  of  estimating  the  magnitude  of  solid  angles 
appears  to  have  been  first  given  by  Albert  Girard  in  his  Invention  Nouvelle 
en  Algebre,  1629 ;  and  it  would  very  naturally  suggest  itself  as  one  of  the 
simplest  applications  of  his  theorem  for  the  spherical  excess.)"  [Compare 
Plane  Trigonometry,  p.  126,  Note  2.] 

Ex.  1.  The  edge  angles  of  a  triedral  angle  are  74°  40',  67°  30',  49°  50' ; 
calculate  its  spherical  degree  measure,  and  its  spherical  measure.  (See  Ex. 
1,  Art.  43.) 

Ex.  2.  The  face  angles  of  a  triedral  angle  are  47°  30',  55°  40',  60°  10' ; 
calculate  its  spherical  degree  measure,  and  its  spherical  measure.  (See  Ex. 
1,  Art.  42.) 

Ex.  3.  Two  face  angles  of  a  triedral  angle  are  64°  24  ,  42°  30',  and  the 
edge  angle  between  their  planes  is  58°  40'  ;  calculate  its  spherical  degree 
measure,  and  its  spherical  measure.  (See  Ex.  1,  Art.  44.) 

Ex.  4.  A  face  angle  of  a  triedral  angle  is  74°  20',  and  the  two  adjacent 
edge  angles  are  67°  30'  and  45°  50'  j  calculate  its  measure.  (See  Ex.  1,  Art. 
45.) 

Ex.  5.  Calculate  the  spherical  degree  measure,  and  the  spherical  measure, 
of  the  solid  angles  corresponding  to  the  spherical  triangles  described  in 
Art.  42,  Exs.  2,  3  ;  Art.  43,  Ex.  2 ;  Art.  44.  Exs.  2,  3 ;  Art.  45,  Ex.  2  j  Art. 
46,  Exs.  2,  3 ;  Art.  47,  Ex.  2.  (See  Ex.,  Art.  56.) 


78  SPHERICAL   TRIGONOMETRY.  [Cn.  VI. 

62.  The  volume  of  a  sphere.  In  some  works  on  solid  geometry 
and  in  books  on  mensuration  it  is  shown  that  the  volume  of  a 
pyramid  is  equal  to  one  third  the  product  .-:J  its  base  and  altitude. 
Now  suppose  that  a  polyedron  (i.e.  a  solid  bounded  by  plane  faces) 
is  circumscribed  about  a  sphere,  each  of  the  faces  of  the  polye- 
dron, accordingly,  touching  the  sphere.  This  polyedron  may  be 
regarded  as  made  up  of  pyramids  which  have  a  common  vertex 
(namely,  the  centre  of  the  sphere),  and  a  common  altitude  (namely, 
the  radius  of  the  sphere),  and  which  have  the  faces  of  the  poly- 
edron as  bases.  Then,  R  being  the  radius  of  the  sphere, 

Vol.  of  polyedron  =  ^  R  x  (sum  of  faces  of  polyedron).     (1) 

If  the  number  of  faces  of  the  polyedron  be  increased  and  the 
area  of  each  face  be  decreased,  then  the  sum  of  the  faces  becomes 
more  nearly  equal  to  the  area  of  the  surface  of  the  sphere,  and 
the  volume  of  the  polyedron  becomes  more  nearly  equal  to  the 
volume  of  the  sphere.  By  increasing  the  number  of  faces  and 
decreasing  the  area  of  each  face,  the  difference  between  the  sum 
of  the  faces  of  the  polyedron  and  the  area  of  the  sphere  can  be 
made  as  small  as  one  please  ;  and,  likewise,  the  difference  between 
the  volume  of  the  polyedron  and  the  volume  of  the  sphere  can  be 
made  as  small  as  one  please.  In  other  words  : 

The  area  of  the  surface  of  the  sphere  is  the  limit  of  the  area  of 
the  surface  of  the  polyedron,  and  the  volume  of  the  sphere  is  the 
limit  of  the  volume  of  the  polyedron,  when  the  faces  of  the  latter 
are  increased  without  limit,  and  each  face  is  made  to  approach 
zero  in  area. 

Hence,  from  (1),  Vol.  of  sphere  =  \  R  x  surface  of  sphere 

(2) 


63.  Definitions.  A  spherical  pyramid  is  a  portion  of  a  sphere 
bounded  by  a  spherical  polygon  and  the  planes  of  the  sides  of  the 
polygon.  The  polygon  is  called  the  base  of  the  pyramid. 

*  For  a  note  concerning  the  measurement  of  the  circle  and  the  sphere  see 
Plane  Trigonometry,  Art.  72,  and  Note  C,  p.  171.  For  the  proofs  of  Archi- 
medes, see  T.  L.  Heath,  The  Works  of  Archimedes  edited  in  modern  notation, 
with  introductory  chapters  (Cambridge,  University  Press),  pp.  39,  41,  93. 


62-64.]  VOLUME  OF  A   SPHERICAL  PYRAMID.  79 

For  example,  in  Fig.  11,  Art.  12,  0-ABCD,  0-ABC,  0-ABD, 
are  spherical  pyramids ;  their  bases  are  ABCD,  ABC,  ABD. 

A  spherical  sector  is  the  portion  of  a  sphere  generated  by  the 
revolution  of  a  sector  of  a  circle  about  any  diameter  of  the  circle 
as  axis.  For  example,  in  Fig.  47,  Art.  53,  when  the  semicircle 
ATB  revolves  about  AB,  each  of  the  circular  sectors  AOL,  LOT, 
LOK,  etc.,  describes  a  spherical  sector. 

A  spherical  segment  is  the  portion  of  a  sphere  bounded  by  two 
parallel  planes  and  the  zone  intercepted  between  them.  (One  of 
the  planes  may  be  tangent  to  the  sphere.) 

64.  Volume  of  a  spherical  pyramid ;  of  a  spherical  sector.  By 
reasoning  analogous  to  that  in  Art.  62,  it  can  be  shown  that,  in  a 
sphere  of  radius  R, 

vol.  of  a  spherical  pyramid  =  ^  R  x  area  of  its  base ; 
vol.  of  a  spherical  sector  =  J  R  x  area  of  its  zone. 
Since  the  area  of  a  zone  of  height  h  =  2  -rrRh  (Art.  53), 
then  vol.  of  spherical  sector  =  f  irl&h. 

Thus  in  Fig.  11,  Art.  12, 

vol.  0-ABCD  =  i  OA  x  area  ABCD; 
in  Fig.  47,  Art.  53, 

vol.  of  sector  described  by  AOL  =  ±  OA  x  area  of  zone  described 
by  arc  AL  =  f  TT R2  -  Al,  and 

rol.  of  sector  described  by  LOT  =  1  OA  x  area  of  zone  described 
by  arc  LT  =  £  irR2  -  It. 

EXAMPLES. 

1.  Find  the  volumes  of  the  spherical  pyramids  whose  bases  are  the  tri- 
angles described  in  Art.  57,  Exs.  1-6. 

2.  Find  the  volumes  of  the  following  spherical  sectors  : 

(a)  The  sector  whose  base  is  a  zone  of  height  2  inches  on  a  sphere  of 
radius  18  inches. 

(6)  The  sector  whose  base  is  a  zone  of  height  3  feet  on  a  sphere  of  radius 
12  feet. 


80 


SPHERICAL    TRIGONOMETRY. 


[Cn.  VI. 


65.   Volume  of  a  spherical  segment.     Let  AB  be  an  arc  of  a  semi- 
circle of  radius  R  having  the  diameter  DD'.     From  A,  B,  draw 
Aa,  Bb,  at  right  angles  to  DD'.     It  is  required  to  find  the  volume 
of  the  spherical  segment  generated  by  the  revo- 
lution of  ABba  about  DD'. 

Let  h  denote  the  height  of  the  segment,  and 
PD  P2>  the  lengths  of  the  perpendiculars  from 
the  centre  0  to  the  parallel  bases  of  the  seg- 
ment. On  making  the  revolution  of  the  semi- 
circle DAD',  it  is  seen  that 

segment   generated   by  ABba  =  cone   generated 
by  BOb  +  spherical  sector  generated 
by  AOB  —  cone  generated  by  A Oa. 

Now,  vol.  cone  generated  by  BOb  =  1 7rr22p2; 

vol.  sector  generated  by  AOB  =  f  irR^h ;       (Art.  64) 

vol.  cone  generated  by  AOa  =  ^-n-Tip^ 
.-.  vol  segment  =  1  TT  (r£p2  -f  2  R~h  -  rfa).  (1) 

NOTE.     The  result  (1)  can  be  reduced  to  various  forms.    For  example, 
since 


then    vol.  segment  =  §*•  R*(pt  -j>0+  |  *ps(R*  -p22)- 


Since 


h=p2—  p\,  then  7i2  =  p22  —  2  j92  px  -f  pi2. 
+  P-V,  and 


On  substituting  the  last  result  in  (3),  expressing  p^  and  p£  in  terms  of 
,  ri,  r2,  and  reducing,  the  following  formula  is  obtained,  viz.  : 


vol.  segment  = 


+  r22  + 


(4) 


65.]  VOLUME  OF  A   SPHERICAL   SEGMENT.  81 


EXAMPLES. 

1.  Show  that  if  (in  Fig.  51)  angle  AOD  =  «,  then  the  volume  of  the 
spherical  sector  generated  by  AOD  is  f  TT R»(\  —  cos  a). 

2.  Show  that  if  angle  AOD  =  «,  then  the  volume  of  the  segment  generated 
by  the  revolution  of  ADa  is  |  wR3  sin*  \  a(l  +  2  cos2 1  a). 

SUGGESTION.     Segment  generated  by  ADa  =  sector  generated  by  AOD  — 
cone  generated  by  AOa. 

3.  Find  the  volume  of  a  spherical  segment,  the  diameters  of  its  ends  being 
10  and  12  inches,  and  its  height  2  inches. 

4.  The  diameters  of  the  ends  of  a  spherical  segment  are  8  and  12  inches, 
and  its  height  is  10  inches.    Find  its  volume. 

N.B.     For  questions  and  exercises  on  Chapter  VI. ,  seepage  108. 


CHAPTER   VII. 

PRACTICAL   APPLICATIONS. 

66.  Geographical  problem.  To  find  the  distance  between  two  places 
and  the  bearing  (i.e.  the  direction)  of  each  from  the  other,  when  their 
latitudes  and  longitudes  are  known.  An  interesting  application  of 
spherical  trigonometry  can  be  made  in  solving  this  problem.  In 
the  following  examples  the  earth,  is  regarded  as  spherical,  and  its 
radius  is  taken  to  be  3960  miles. 

EXAMPLES. 

1.  Find  the  shortest  distance  along  the  earth's  surface  hetween  Baltimore 
(lat.  39°  17'  N.,  long.  76°  37'  W.)  and  Cape  Town  (lat.  33°  56'  S.,  long. 
18°26'E.). 

In  Fig.  52  B  and  C  represent  Baltimore  and  Cape  Town  ;  EQ  is  the  earth's 
equator ;  NGS,  NBS,  NCS  are  the  meridians  of 
Greenwich,  Baltimore,  and  Cape  Town  respec- 
tively ;  BC  is  the  great  circle  arc  whose  length  is 
required. 

In  the  spherical  triangle  BNC,  NB,  JVC,  and 
BNC  are  known.     For 

NB  =  90°  -  BL  =  90°  -  39°  17'  =  50°  43' 
NO  =  90°  +  TO  =  90°  +  33°  56'  =  123°  56' 
BNC  =  BNG  +  GNC  =  76°  37'  +  18°  26'  =  95°  3' 

Hence,  BC  can  be  determined  in  degrees  by  Art.  44  ;  then,  the  radius 
of  the  sphere  being  given,  BC  can  be  determined  in  miles.  The  angles  NBC, 
NCB,  can  also  be  found. 

Answers :  BC  =  (65°  47'  48")  =  4685.8  miles  ;  NBC  =  115°  1'  35'  ;  NCB 
=  57°  42'  23". 

NOTE  1.  The  bearing  of  one  place  from  a  second  place  is  the  angle  which 
the  great  circle  arc  joining  the  two  places  makes  with  the  meridian  of  the 
second  place.  Thus,  in  Fig.  52  the  bearing  of  Cape  Town  from  Baltimore  is 
the  angle  NBC,  and  the  bearing  of  Baltimore  from  Cape  Town  is  NCB. 

82 


66.]  PRACTICAL  APPLICATIONS.  83 

Since  NBC  =  115°  lf  35"  the  ship  sets  out  from  Baltimore  on  a  course 
S.  64^  58''  25"  E.;  since  NCB=bl°  42'  23"  the  ship  approaches  Cape  Town 
on  a!  course  S.  57°  42'  23"  E. 

NOTE  2.  A  ship  that  sails  on  a  great  circle  (excepting  the  equator  or  a 
meridian)  must  be  continually  changing  her  course. 

2.  Find  the  latitude  of  the  place  where  BC  crosses  the  meridian  15°  W.  ; 
also  find  the  bearing  of  Cape  Town  from  this  place. 

3.  If  a  vessel  sails  from  Baltimore  and  keeps  constantly  on  the  course 
(see  Ex.  1)  S.  64°  58'  25"  E.  (i.e.  crosses  every  meridian  at  the  angle  64°  58' 
25"),  will  she  arrive  at  Cape  Town?     [Answer.     No.] 

4.  What  path  will  the  vessel  in  Ex.  3  make  on  the  sea  ?    Answer.     A 
path  which  is  a  spiral  going  round  and  round  the  earth  and  gradually 
approaching  the  south  pole.     This  path  is  called  the  loxodrome,  or  rhumb 
line. 

5.  If  a  person  leaves  Boston,  Mass.  (lat.  42°  21'  N.,  long.  71°  4'  W.),  start- 
ing due  east,  and  keeps  on  a  great  circle  :  (a)  Where  will  he  be  after  he  has 
passed  over  an  arc  of  90°,  and  in  what  direction  will  he  be  going  ?    (&)  Where 
will  he  be  after  he  has  passed  over  an  arc  of  180°,  and  in  what  direction  will 
he  be  going  ?     (c)  Where  will  he  be  after  he  has  passed  over  an  arc  of  270°, 
and  in  what  direction  will  he  be  going  ?      [Solve  this  example :    (1)  by 
spherical  geometry  ;    (2)  by  spherical  trigonometry.] 

6.  What  is  the  distance  from  New  York  (40°  43'  N. ,  74°  0'  W. )  to  Liverpool 
(53°  24'  N.,  3°  4'  W.)?    Find  the  bearing  of  each  place  from  the  other.     In 
what  latitude  will  a  steamer  sailing  on  a  great  circle  from  New  York  to  Liver- 

)1  cross  the  meridian  of  50°  W. ;  and  what  will  be  her  course  at  that  point? 

N.B.     Check  the  results  in  the  following  exercises  : 

7.  Find  the  distance  and  bearing  of  Liverpool  from  Montreal  (45°  30'  N., 
73°  33'  W.). 

8.  Find  the  distance  and  bearing  of  Liverpool  from  Halifax,  N.  S.  (44° 
40' N.,  63°  35'  W.). 

9.  Find  the  distance  and  bearing  of  Santiago  de  Cuba  (20°  N.,  75°  50'  W.) 
from  Rio  de  Janeiro  (22°  54'  S.,  43°  8'  W.). 

10.  Find  the   distance  and  bearing  of  San  Francisco  (37°  47'  55"  N., 
122°  24'  32"  W.)  from  New  York. 

11.  Find  the  distance  of  Victoria,  B.  a  (48°  25'  N.,  123°  23'  W.)  from 
Sydney,  N.  S.  W.   (33°  52'  S.,  151°  13'  E.)  ;  and  the  bearing  of  each  place 
from  the  other. 


84  SPHERICAL    TRIGONOMETRY.  [Cn.  VII. 

12.  Find  the  distances  between  the  following  places :     (a)  San  Francisco 
and  Honolulu ;    (&)   Cape   Town  and   Cairo ;    (c)    Honolulu  and   Manila ; 
(d)  Victoria,  B.  C.,  and  Tokio. 

13.  Find  the  distances  between  other  places,  and  their  bearings  from  each 
other. 

^ 

APPLICATIONS   TO   ASTRONOMY. 

N.B.  In  connection  with  his  study  of  the  following  articles  the 
student  should  consult  some  elementary  text-book  on  astronomy.  The 
numerical  examples  given  here  will  supplement  his  outside  reading  on 
spherical  astronomy. 

67.  One  of  the  most  important  applications  of  spherical  trigo- 
nometry is  to  astronomy.      Trigonometry  was  invented  to  aid 
astronomy,  and  for  centuries  was  studied  as  an  adjunct  of  the 
latter  subject.     (See  Plane  Trigonometry,  pp.  165,  166.)     A  few  of 
the  simplest  problems  of  spherical  astronomy  are  introduced  in 
Arts.  73,  74.     In  order  to  understand  these  problems  a  clear  con- 
ception of  a  few  astronomical  terms  and  principles  is  necessary. 
These  terms  are  explained  in  Arts.  68-72. 

68.  The  celestial  sphere.     To  a  person  on  the  surface  of  the 
earth,  the  sky  above  is  like  a  great  hemispherical  bowl  with  him- 
self at  the  centre.     The  stars  seem  to  move  from  east  to  west 
across  the  spherical  sky  in  parallel  circles  whose  axis  is  the  earth's 
polar  axis  prolonged.      Each  star  makes  a  complete  revolution 
about  this  axis  in  23  hours  56  minutes  ordinary  clock  time.     The 
stars  appear  never  to  change  their  positions  with  reference  to  one 
another,  being  in  this  respect  like  places  on  the  earth's  surface.* 
Another  way  of  describing  the  relations  of  the  earth  and  the 
enveloping  sky,  is  to  say  that  the  whole  sky  is  turning,  like  an 
immense  crystal  sphere,  about  an  axis  which  is  the  earth's  polar 
axis  prolonged,  the  motion  being  from  east  to  west.      The  stars 
keep  the  same  positions  with  respect  to  one  another,  and,  accord- 
ingly, appear  to  be  attached  to  the  surface  of  the  sphere.     As  the 
sphere  turns,  the  stars  fixed  in  it  appear  to  trace  parallel  circles 

*  The  positions  of  some  of  the  stars  suffer  a  very  slight  change  which  is  per- 
ceptible in  the  course  of  centuries. 


67-68.]  APPLICATIONS   TO  ASTRONOMY.  85 

about  the  axis.  The  sphere  turns  completely  in  23  hours  56 
minutes  ordinary  clock  time.*  The  stars  all  seem  to  be  at  the 
same  distance  from  the  observer  because  his  eyes  can  judge  their 
directions  only,  and  not  their  distances. 

The  following  considerations  will  show  that  it  is  natural  enough  for  an 
observer  on  the  earth  to  think  that  he  is  always  at  the  centre  of  the  sphere  on 
which  the  stars  appear  to  be.  When  a  person  changes  his  position,  the  direc- 
tion of  an  object  at  which  he  is  looking  changes  also,  unless  he  moves  directly 
towards  or  away  from  the  object.  For  instance,  from  a  certain  point  a  tree 
may  be  in  an  easterly  direction,  and  when  the  observer  moves  a  little  way 
the  tree  may  be  in  a  southeasterly  direction.  Moreover,  the  further  away  an 
object  is,  the  less  will  be  the  change  in  its  direction  caused  by  any  particular 
change  in  the  observer's  position.  Thus,  if  a  person  is  near  a  tree,  a  few  steps 
on  his  part  may  change  the  direction  of  the  tree  from  east  to  southeast,  but  if  he 
is  five  miles  from  the  tree,  an  equal  number  of  steps  taken  by  him  will  make 
very  little  difference  in  the  direction  of  the  tree.  Now  the  earth's  mean  distance 
from  the  sun  is  about  93,000,000  miles.  Hence,  an  observer  who  now  looks  at 
the  stars  from  a  certain  position,  in  about  six  months  from  now  will  look  at 
them  from  a  point  186,000,000  miles  distant  from  his  present  position,  t  As- 
tronomers have  succeeded  in  a  few  instances  in  determining  the  distances  of 
the  stars  from  the  earth,  t  It  has  been  found  that  the  nearest  star  yet  known, 
Alpha  Centauri,  is  so  far  away  that  the  change  in  its  direction  from  the  centre 
df  the  earth,  due  to  the  change  of  position  of  186,000,000  miles  on  the  part  of 
the  earth,  is  less  than  the  change  in  the  direction  of  an  object  3£  miles  away 
when  the  observer  moves  his  head  a  couple  of  inches  at  right  angles  to  the 
line  of  sight.  This  being  so  in  the  case  of  the  sun's  nearest  stellar  neighbour,  it 
is  natural  for  an  observer  on  the  earth  to  think  that  he  is  always  at  the  centre 
of  the  great  sphere  on  which  the  stars  appear  to  be  ;  and  it  is  perfectly  proper 

*  The  student  probably  knows  that  the  apparent  turning  of  the  spherical 
sky  from  east  to  west  about  an  axis  which  is  the  earth's  polar  axis  prolonged, 
is  really  due  to  the  rotation  of  the  earth  in  an  opposite  direction.  The 
observer  is  not  conscious  of  any  motion  of  the,  earth,  and  thinks  that  the  sky 
with  its  bright  points  is  revolving  about  the  earth  from  east  to  west,  while  all 
the  time  the  sky  is  motionless,  and  the  earth  is  turning  under  it  from  west  to 
east.  Just  as  to  a  person  in  a  swiftly  moving  train  the  objects  outside  seem 
to  be  rushing  by  him  while  the  train  appears  to  be  at  rest. 

t  This,  moreover,  does  not  take  any  account  of  the  motion  of  the  sun  with 
his  system  through  space. 

J  The  first  stellar  distance  determined  was  that  of  61  Cygni  by  Friedrich 
Wilhelm  Bessel  (1784-1846),  one  of  the  greatest  of  German  astronomers,  in 
1838.  Si-nce  then  the  distances  of  about  100  stars  have  been  measured ;  about 
50  of  these  distances  are  regarded  as  reliably  determined. 


86  SPHERICAL   TRIGONOMETRY  [Cn.  VII. 

for  him  to  act  in  accordance  with  this  notion  when  he  makes  astronomical 
observations  and  deductions.* 

Thejsphere  on  which  the  stars  appear  to  move  in  parallel  cir- 
cles, or,  what  comes  to  the  same  thing,  the  sphere  which  appears 
to  have  the  stars  attached  to  it  and  to  revolve  about  the  earth's 
polar  axis  prolonged,  is  called  the  celestial  sphere. 

69.  Points  and  lines  of  reference  on  the  celestial  sphere.  There 
will  now  be  shown  some  methods  for  indicating  the  positions  of 
the  heavenly  bodies  on  the  celestial  sphere  —  their  positions  with 
respect  to  the  observer  and  their  positions  with  respect  to  one  another. 

The  positions  of  places  on  the  terrestrial  sphere  are  described 
by  means  of  certain  points  and  great  circles  on  the  sphere.  There 
are  various  pairs  of  circles  which  are  used  for  reference;  for 
example,  the  equator  (whose  poles  are  the  north  and  south  poles 
of  the  earth)  and  the  meridian  passing  through  the  Koyal  Observa- 
tory at  Greenwich,  the  equator  and  the  meridian  passing  through 
the  observatory  at  Washington,  etc.  It  will  be  observed  that  in 
each  case  the  reference  circles  are  at  right  angles  to  each  other,  and, 
accordingly,  each  of  them  passes  through  the  poles  of  the  other. 

In  an  analogous  manner  the  positions  of  bodies  on  the  celestial 
sphere  are  described  by  means  of,  or  by  reference  to,  certain  points 
and  great  circles  on  that  sphere.  There  are  four  different  systems 
of  circles  of  reference.  As  in  the  case  of  the  terrestrial  sphere, 
each  system  consists  of  two  circles,  each  of  which  passes  through 
the  pole  of  the  other,  and,  accordingly,  is  at  right  angles  to  the 
other.  Two  of  these  systems  are  described  in  Arts.  70, 71,  a  third 
in  Art.  76,  and  the  fourth  in  Art.  77.  A  point  which  will  be  referred 
to  in  these  systems  is  the  north  celestial  pole.  This  is  the  point 
where  the  earth's  axis,  if  prolonged,  would  pierce  the  celestial 
sphere.  It  is  near  the  pole  star,  being  about  1^°  from  it. 

***...  imagine  the  entire  solar  system  as  represented  by  a  tiny  circle  the 
size  of  the  dot  over  this  letter  i."  (Neptune  the  outermost  planet  known  of 
the  solar  system  is  2700  millions  of  miles  from  the  sun  ;  i.e.  30  times  as  far 
as  the  earth.)  "  Even  the  sun  itself,  on  this  exceedingly  reduced  scale,  could 
not  be  detected  with  the  most  powerful  microscope  ever  made.  But  on  the 
same  scale  the  vast  circle  centred  at  the  sun  and  reaching  to  Alpha  Centauri 
would  be  represented  by  the  largest  circle  which  could  be  drawn  on  the  floor 
of  a  room  10  feet  square."  (Todd,  New  Astronomy,  p.  438.) 


69-70.]          CIRCLES   ON  THE  CELESTIAL   SPHERE.  87 

70.  The  horizon  system :  Positions  described  by  altitude  and  azi- 
muth. For  any  place  on  the  earth's  surface,  the  point  at  which 
the  plumb  line  extended  upwards  meets  the  celestial  sphere  is 
called  the  zenith ;  the  diametrically  opposite  point  is  called  the 
nadir.  If  a  plane  perpendicular  to  the  plumb  line  be  passed  either 
immediately  beneath  the  observer's  feet,  or  through  the  centre  of 
the  earth,  about  4000  miles  below  him,  then  the  intersection  of  this 
plane  with  the  celestial  sphere  is  called  the  horizon.  (Since  the 
earth  is  so  small  and  so  far  away  from  even  the  nearest  star,  two 
parallel  planes  4000  miles  apart  and  passing  through  the  earth 
will  appear,  to  a  terrestrial  observer,  to  intersect  the  celestial 
sphere  in  the  same  great  circle.) 

Great  circles  passing  through  the  zenith  are  perpendicular  to  the 
horizon ;  they  are  called  vertical  z  (Zenith) 

circles.  The  north  point  of  the  horizon 
is  the  point  which  is  directly  north 
from  the  observer.  It  is  where  the 
vertical  circle  passing  through  the  Soutt^( 
north  pole  intersects  the  horizon. 
This  circle  which  passes  through  the 
zenith  and  the  pole  is  called  the  me- 
ridian of  the  observer.  The  horizon 
and  the  meridian  are  the  reference  circles  in  the  horizon  system. 

The  altitude  (denoted  by  Ji)  of  a  heavenly  body  is  its  angular 
distance  above  the  horizon.  Thus  the  altitude  of  M  (Fig.  53,  in 
which  E  is  the  earth  and  Z  the  zenith  of  the  place  of  observation) 
is  Mm.  The  altitude  of  the  zenith  is  90°.  The  distance  of  a  star 
from  the  zenith  is  called  its  zenith  distance  ;  this  is  obviously  the 
complement  of  the  altitude. 

The  azimuth  (denoted  by  A)  of  a  heavenly  body  is  the  angle 
between  its  vertical  circle  and  the  meridian.  This  angle  is 
measured  usually  along  the  horizon  from  the  south  point  in  the 
direction  of  the  west  point,  to  the  foot  of  the  star's  vertical  circle. 
Thus  in  Fig.  53  the  azimuth  of  M  is  180°  +  NZm,  which  is  measured 
by  the  arc  180°  +  Nm  on  the  horizon. 

NOTE.  Any  two  points  on  the  earth's  surface  have  different  zeniths. 
Hence,  the  above  system  of  describing  positions  on  the  celestial  sphere  is 
peculiarly  local.  Moreover,  a  star  rises  in  the  eastern  part  of  the  horizon 


88 


SPHERICAL    TRIGONOMETRY. 


[CH.  VII. 


(altitude  zero),  mounts  higher  in  the  sky  until  it  reaches  the  observer's  merid- 
ian, then  sinks  towards,  and  sets  in,  the  west ;  it  is,  accordingly,  continually 
changing  its  altitude  and  azimuth. 

71.  The  equator  system:  Positions  described  by  declination  and 
hour  angle.  The  north  celestial  pole  is  the  principal  point  of  this 
system.  The  celestial  equator  is  the  great  circle  of  which  that  point 
is  the  pole ;  it  is  evidently  the  projection  of  the  earth's  equator 
upon  the^celestial  sphere.  The  celestial  equator  and  the  meridian  of 
the  observer  are  the  reference  circles  in 
the  system  now  being  described.  In 
Fig.  54,  P  is  the  north  celestial  pole,  S 
the  south  celestial  pole,  EQ,  the  celestial 
equator ;  also,  HR  is  the  horizon  and  Z 
the  zenith  for  some  particular  place  on 
the  earth's  surface.  As  said,  in  Art. 
68,  the  stars  move  in  parallel  circles 
whose  axis  is  PS ;  these  circles  are,  accord- 
ingly, parallel  to  the  equator  EQ,.  The 
angular  distance  of  a  star  from  the  equator  is  called  the  declination 
(denoted  by  D  or  8)  of  the  star ;  north  (or  +)  declination  when  the 
star  is  north  of  the  equator,  and  south  (or  — )  declination  when  the 
star  is  south.  Thus  the  declination  of  S^  is  $3<V  The  angular  dis- 
tance of  a  star  from  the  north  pole  is  called  its  north  polar  distance  ; 
this  is  evidently  the  complement  of  the  star's  declination.'* 

In  24  (sidereal)  hours  a  star  appears  to  make  a  complete  revolu- 
tion (i.e.  to  pass  over  360°)  about  the  celestial  polar  axis ;  hence, 
the  star  passes  over  15°  in  1  hour.-f  The  great  circles  passing 
through  the  poles  are  called  hour  circles.  Thus  PSSS  is  the  hour 
circle  of  $3.  The  hour  angle  (denoted  by  //.  JL.)  of  a  star  is  the  angle 
between  the  meridian  of  the  observer  and  the  hour  circle  of  the 


Fia.  54 


*  The  declination  of  the  stars  change  by  an  exceedingly  small  amount  in 
the  course  of  a  year. 

t  The  interval  of  time  between  two  successive  passages  of  the  observer's 
meridian  by  the  sun  (i.e.  from  noon  to  noon)  is  about  4  minutes  longer  than 
the  interval  of  time  between  two  successive  passages  of  the  meridian  by  any 
particular  star.  (This  difference  is  due  to  the  yearly  revolution  of  the  earth 
about  the  sun.  See  text-books  on  astronomy.)  The  second  interval  is 
called  a  sidereal  day  ;  it  is  divided  into  24  sidereal  hours. 


71-72.] 


DECLINATION  AND  HOUR  ANGLE. 


star.  This  angle  is  measured  towards  the  west.  Thus,  suppose 
that  a  star  is  on  the  meridian  at  S4 ;  its  hour  angle  is  then  zero. 
Twelve  hours  later  the  star  will  be  at  S0,  and  will  have  an  hour 
angle  180°.  After  a  while  it  will  be  at  Sl}  just  rising  above  the 
horizon,  and  its  hour  angle  will  be  180°  -|-  S0PSl ;  later  it  will  be 
at  $3,  having  the  hour  angle  180°  +  SoPS3 ;  later  still  it  will  be  on 
the  meridian  at  S4,  and  its  hour  angle  will  be  zero  again.  The  hour 
angle  is  usually  reckoned  in  hours  from  1  to  24,  1  hour  being  equal 
to  15  degrees.  Thus,  when  the  star  is  at  $0  its  hour  angle  is  12  h. 
The  hour  angle  of  a  star  is  partly  local;  for  only  places  on  the  same 
meridian  of  longitude  have  the  same  celestial  meridian.  More- 
over, the  hour  angle  of  a  star  is  continually  changing,  and  its 
magnitude  depends  upon  the  time  of  observation.  In  Arts.  76, 
77,  the  positions  of  stars  are  described  in  terms  which  are  inde- 
pendent of  the  time  and  place  of  observation. 

In  Arts.  73,  74,  75,  the  astronomical  ideas  so  far  obtained,  are 
used  in  the  solution  of  two  simple  problems. 

72.  The  altitude  of  the  pole  is  equal  to  the  latitude  of  the  place  of 
observation.  This  theorem,  which  is  necessary  in  Arts.  73, 74,  is  the 
fundamental  and  most  important  theorem  of  spherical  astronomy. 

In  Fig.  55,  C  represents  the 
centre  of  the  earth,  P  its  north 
pole,  and  EQ  its  equator ;  0  is  the 
place  of  observation,  say  some  place 
in  the  northern  hemisphere,  Z  is  its 
zenith  and  HR  its  horizon ;  CPPl  is 
the  celestial  polar  axis,  Pl  being  the 
north  celestial  pole.  Draw  OP2 
parallel  to  CPlt  P2  being  on  the 
celestial  sphere.  The  angle  ROP2 
is  the  altitude  of  the  pole  at  0, 
since  (see  Arts.  68,  70)  P1  and  P2  are  in  the  same  direction  from  0. 

The  latitude  of  a  place  is  equal  to  the  angle  between  the  plumb 
line  and  the  plane  of  the  equator.  Thus,  the  latitude  of  0  is  equal 
to  OCE.  Since  OR  and  OP2  are  respectively  perpendicular  to 
CZ  and  CE,  the  angle  jROP2=  OCE;  that  is,  the  altitude  of  the 
pole  as  observed  at  0  is  equal  to  the  latitude  of  0. 


Fia.  55 


00  SPHERICAL   TRIGONOMETRY.  [Cn.VII. 

73.  The  time  of  day  can  be  determined  at  any  place  ivhose  latitude 
is  known,  if  the  declination  and  the  altitude  of  the  sun  at  that  time 
and  place  are  also  known. 

NOTE  1.  The  sun,  unlike  the  stars,  changes  in  declination  from  23|°  south 
(about  Dec.  22)  to  231°  north  (about  June  21),  and  then  returns  south.  Its 
declination  is  zero,  that  is,  it  is  on  the  celestial  equator,  about  March  20  and 
Sept.  22.  This  change  in  declination  is  due  to  the  revolution  of  the  earth 
about  the  sun,  and  to  the  fact  that  the  plane  of  the  earth's  equator  is  inclined 
about  23|°  to  the  plane  of  its  orbit  about  the  sun.  The  latter  plane  is  called 
the  plane  of  the  ecliptic.  The  declination  of  the  sun  is  given  for  each  day 
of  the  year  in  the  Nautical  Almanac.  The  altitude  of  the  sun  can  be  observed 
with  a  sextant. 

NOTE  2.  The  student  should  consult  a  text-book  on  astronomy  for  an 
account  of  the  special  precautions  and  corrections  necessary  in  connection 
with  this  and  similar  astronomical  problems. 

In  Fig.  56,  P  is  the  north  celestial 
pole,  EQ  the  celestial  equator,  /S  the 
sun,  and  S0SSn  is  the  small  circle  on 
which  the  sun  is  moving  at  the  given 
time;  Z  is  the  zenith,  and  HR  the 
horizon,  of  the  place  of  observation; 
ZSM  is  .  the  sun's  vertical  circle,  and 
PSN  is  its  hour  circle. 

^  ig  midnight  when  the  sun  is  at  SQ, 
and  noon  when  the  sun  is  at  Sn.  From 

noon  to  noon  is  24  hours.  Hence,  to  find  the  time  when  the  sun  is 
at  S,  determine  the  angle  ZPS  in  hours  (15°  =  1  h.)  ;  subtract  the 
number  of  hours  from  12,  if  it  is  forenoon;  and  add,  if  it  is 
afternoon. 

Let  I,  h,  D,  respectively,  denote  the  latitude  of  the  place,  and 
the  altitude  and  declination  of  the  sun. 

Then  PR  =  I  (Art.  72),  SM=  h,  SN=  D. 

In  ZPS,  whose  vertices  are  the  sun,  zenith,  and  pole, 


Hence,  the  angle  ZPS  can  be  found. 


73-74.]  TO  FIND   THE  TIME  OF  DAY.  91 

EXAMPLES. 

1.  In  New  York  (lat.  40°  43'  N.)    the  sun's  altitude   is   observed  to  be 
30° 40'.     What  is  the  time  of  day,  given  that  the  sun's  declination  is  10°  N., 
and  the  observation  is  made  in  the  forenoon  ? 

2.  In  Montreal  (lat.  45°  30' N.)  at  an  afternoon  observation  the  sun's 
altitude  is  26°  30'.     Find  the  time  of  day,  given  that  the  sun's  declination 

is  8°S. 

3.  In  London  (lat.  51°  30'48"N.)  at  an  afternoon  observation  the  sun's 
altitude  is  15°  40'.     Find  the  time  of  day,  given  that  the  sun's  declination 
is  12°  S. 

4.  As  in  Ex.  2,  given  that  the  sun's  declination  is  18°  N. 

5.  As  in  Ex.  3,  given  that  the  sun's  declination  is  22°  N. 

6.  As  in  Ex.  1,  given  that  the  sun's  declination  is  10°  S. 

74.  To  find  the  time  of  sunrise  at  any  place  whose  latitude  is 
knoivn,  when  the  sun's  declination  is  also  known.  This  is  a  special 
case  of  the  preceding  problem ;  for  at  sunrise  the  sun  is  on  the 
horizon  and  its  altitude  is  zero.  The  problem  can  also  be  solved 
by  means  of  the  triangle  RPSl  (instead 
of  ZPS^  which  is  employed  in  Art  73). 
For,  in 


=  90°  -  D,  PR  =  I,  PRSt  =  90°. 


=  tan  I  tan  D. 


The  angle  RPSl  (i.e.  SoPS^  reduced 
to  hours,  gives  the  time  of  sunrise  (after 

midnight).  If  ZPSl  is  found,  then  ZPS^  reduced  to  hours  and 
subtracted  from  12  (noon),  gives  the  time  of  sunrise.  The  time  of 
sunset  is  about  as  many  hours  after  noon  as.  the  tirno  of  sunrise 
is  before  it. 

In  Fig.  57  the  sun  is  north  of  the  equator.  When  the  sun  is 
south  of  the  equator,  PSl  =  90°  +  D,  and  RPSl  >  90°  for  places 
in  the  northern  hemisphere.  The  student  can  make  the  figure 
and  investigate  this  case,  and  also  the  case  in  which  the  place  is 
in  the  southern  hemisphere. 


92  SPHERICAL    TEIGONOMETEY.  [Cn.  VII. 


EXAMPLES. 

Find  the  approximate  time  of  sunrise  at  a  place  in  latitude  Z,  when  the 
sun's  declination  is  Z>,  in  the  following  cases : 

1.  I  =  40°  43'  N.  (latitude  of  New  York),  D  equal  to  :  (a)  4°  30'  N.  (about 
April  1);  (6)  15°  10' N.  (about  May  1);  (c)  23°  N.  (about  June  10)  ;'(<*)  5°N. 
(about  Sept.  10);  (e)  6°  S.  (about  Oct.  8);  (/)  15°  S.  (about  Nov.  3);  (g)  23°  S. 

2.  I  =  51°  30'  48" N.  (latitude  of  London),  D  as  in  Ex.  1. 

3.  I  =  60°  N.  (latitude  of  St.  Petersburg),  D  as  in  Ex.  1. 

4.  I  =  70°  40'  7"  N.  (latitude  of  Hammerfest,  Norway,  D  as  in  Ex.  1. 

5.  I  =  29°  58'  N.  (latitude  of  New  Orleans),  D  as  in  Ex.  1. 

6.  I  =  33°  52'  S.  (latitude  of  Sydney,  N.  S.  W.),  D  as  in  Ex.  1. 

7.  Find  the  approximate  time  of  sunrise  for  other  days  and  places. 

75.  Theorem.     If  the  latitude  of  the  place  of  observation  is  known, 
then  the  declination  and  hour  angle  of  a  star  can  be  determined  from 
its  altitude  and  azimuth,  and  vice  versa.     For,  in  the  triangle  ZPS 
(Fig.  56),  ZP  =  90°  -  I,   SP=  90°  -  D,  SZ=  90°  -  h,   SPZ  = 
360°  -  H.  A.,  PZS  =  A  —  180°.     Hence,  if  the  latitude  and  any 
two  of   the   four  quantities,  viz.,  altitude,  azimuth,  declination, 
hour  angle,  be  known,  then  the  remaining  two  can  be  found  by 
solving  the  triangle  SPZ. 

76.  The  equator  system:    Positions  described  by  declination  and 
right  ascension.     In  the  system  in  Art.  71  the  circles  of  reference 
were  the  equator  and  the  meridian  of  the  observer.    In  the  system 
in  this  article  the  circles  of  reference  are  the  equator  and  the  circle 

passinq  throuqh  the  celestial  poles  and  the 

P  (Pole)        1 

vernal  equinox.     The  vernal  equinox  is  one 

of  the  points  where  the  ecliptic  intersects 
\      the  equator ;  namely,  the  point  where  the 
sun,  in  its  (apparent)  yearly  path  among 
the  stars,  crosses  the  equator  in  spring. 
(See  text-book  on  astronomy.)    This  point 
may  be  called  'the  Greenwich  of  the  celestial 
58  sphere.   (  The  ecliptic  is  the  proj  ection  of  the 

plane  of  the  earth's  orbit  on  the  celestial 

sphere.     The  plane  of  the  equator  and  the  plane  of  the  ecliptic  are 
inclined  to  each  other  at  an  angle  of  23 -J-°.     See  Art.  73,  Note  1.) 


75-77.]          DECLINATION  AND  RIGHT  ASCENSION.  93 

The  right  ascension  (denoted  by  E.A.)  of  a  heavenly  body  is 
the  angle  at  the  north  celestial  pole  between  the  hour  circle  of 
the  body  and  the  hour  circle  of  the  vernal  equinox.  This  angle 
is  measured  from  the  latter  circle  towards  the  east,  from  0°  to  360° 
or  1  h.  to  24  h.  ;  it  may  be  measured  by  the  arc  intercepted  on  the 
equator.  Declination  has  been  defined  in  Art.  71. 

In  Fig.  58,  P  is  the  north  celestial  pole,  E2Q  the  equator,  E-f) 
the  ecliptic,  and  V  the  vernal  equinox.  If  S  is  any  star,  then 
for  S  D  =  $M)  and  R  A  =  angle  VPM=  arc 


77.  The  ecliptic  system  :  Positions  described'  by  latitude  and  longi- 
tude. In  this  system  the  point  and  circles  of  reference  are  the 
pole  of  the  ecliptic,  the  ecliptic,  and  the  great  circle  passing  through 
the  pole  of  the  ecliptic  and  the  vernal  equinox.  The  latitude  of  a 
star  is  its  angular  (or  arcual)  distance 
from  the  ecliptic;  its  longitude  is  the 
angle  at  the  pole  of  the  ecliptic  between 
the  circle  passing  through  this  pole  and 
the  vernal  equinox  and  the  circle  passing 
through  this  pole  and  the  star.  This 
angle  may  be  measured  by  the  arc  inter- 
cepted on  the  ecliptic.  It  is  always 
measured  towards  the  east  from  the  vernal  FI(J  r 

equinox. 

In  Fig.  59,  K  is  the  pole  of  the  ecliptic,  E±C  the  ecliptic,  P  the 
pole  of  the  equator,  E2Q  the  equator,  and  V  the  vernal  equinox. 
If  S  is  any  star,  then 

latitude  of  S  =  SM9  longitude  of  S  =  VKM=  VM. 

When  the  latitude  and  longitude  of  a  star  are  known,  its  declination 
and  right  ascension  can  be  found,  and  vice  versa.  For,  in  the  tri- 
angle KPS  (the  triangle  whose  vertices  are  the  star  and  the  poles 
of  the  equator  and  the  ecliptic),  KP=23±°  (since  QFC'  =  23f>), 
KS  =  90°-  lat.,  SKP  =  90°-  long.,  SP  =  90°-  D;  SPK=  VPK 
-VPS  =  90°-  (360°  -  K.A.),  if  S  is  west  of  VP  ;  SPK  =  90°  + 
R.A.,  if  S  is  east  of  VP.  If  any  two  of  these  be  known  besides 
KP,  the  remaining  two  can  be  found  by  solving  KPS. 

N.B.    Questions  and  exercises  on  Chapter  VII.  will  be  found  at  page  109. 


APPENDIX. 


NOTE  A. 


ON  THE   FUNDAMENTAL  FORMULAS   OF   SPHERICAL 
TRIGONOMETRY. 

1.  The  relations  between  the  sides  and  angles  of  a  right-angled  spherical 
triangle  were  obtained  in  Art.  26.     The  law  of  sines  and  the  law  of  cosines 
(Art.  36)  for  any  spherical  triangle  have  been  derived  by  means  of  these 
relations.    (See  Note  1,  Art.  86.)     These  two  laws  can  also  be  derived  directly 
by  geometry  -}  this  is  done  in  Arts.  2,  3,  below.     Moreover,  the  law  of  sines 
can  be  derived  analytically  from  the  law  of  cosines,  as  shown  in  Art.  4.     In 
Art.  5  it  is  shown  how  the  relations  for  right-angled  triangles  can  be  derived 
from  these  two  laws.     Other  relations  between  the  parts  of  a  spherical  tri- 
angle have  been  referred  to  in  Art.  40 ;  these  relations  can  also  be  deduced 
by  means  of  the  law  of  cosines  and  the  law  of  sines.     The  law  of  cosines  is, 
accordingly,  the  fundamental  and  most  important  formula  in  spherical  trigo- 
nometry. 

2.  Direct  geometrical  derivation  of  the  law  of  cosines.     Let  0-ABC 
be  a  triedral  angle,  and  ABC  be  the  corresponding  spherical  triangle  on 
a  sphere  of  radius  OA.     It  is  required  to 

find  the  cosine  of  the  face  angle  COB,  or, 
what  is  the  same  thing,  the  cosine  of  the 
side  CB. 

In  OA  take  any  point  P,  and  through 
P  pass  a  plane  MPN  at  right  angles  to 
the  line  OA.  Then  OPN  and  0PM  are 
right  angles,  and  angle  MPN  =  angle  A. 
Also,  the  measures  (in  degrees)  of  the 
sides  AB,  BC,  CA,  are  the  same  as  the 
measures  of  the  face  angles  COB,  BOA,  AOC,  respectively. 

In  MPN,        MW2  =  MP"  +  PN2  -  2  MP .  PN  cos  MPN',  (1) 

in  M ON,  MN2  =  MO2  +  ON'2  -  2  M 0  •  ON  cos  M ON.  (2) 

95 


96 


SPHERICAL    TEIGONOMETET. 


Hence,  on  equating  these  values  of  MN2  and  transposing, 
2  MO  -  ON  cos  MON  =  I/O2  -  MP2  +  ON2  -  PN2  +  2  MP  -  PNcos  MPN. 


Now    OM*  -  MP*  =  OP,    and    ON*  -  PN    =  OP,   since    OP3/  and 
OP.V  are  right  angles. 

/.  2  MO  -  ON  cos  MON=  2  OP2  +  2  HP  •  PNcos  MPN. 


<.«. 


cos  a  =  cos  &  cos  c  +  sin  &  sin  c  cos  ^1. 


(3) 


Like  formulas  for  cos  6,  cos  c,  can  be  derived  in  a  similar  manner  ;  they 
can  also  be  written  immediately,  on  paying  regard  to  the  symmetry  in  (3). 
The  formulas  for  cos  A,  cos  JS,  and  cos  <7,  can  be  derived  by  means  of  the 
polar  triangle,  as  done  in  Art.  36,  (7. 


EXERCISES. 

1.  Make  the  figure  and  derive  the  law  of  cosines  :  («)  when  P  is  taken 
at  A  ;  (6)  when  P  is  taken  in  OA  produced  towards  A. 

2.  Derive,  the  formula  for  cos  6  geometrically.     (Take  any  point  in  OB, 
and  through  this  point  pass  a  plane  at  right  angles  to  OB.  ) 

3.  Derive  the  formula  for  cos  c  geometrically 

3.   Direct  geometrical   derivation  of  the  law  of  sines.       Let  0-ABC 
be  a  triedral'  angle,  and  ABC  be  the  corresponding  spherical  triangle  on 

a  sphere  of  radius  OA.  % 

In  OC  take  any  point  P,  and  draw  PM 
at  right  angles  to  the  plane  AOB,  and  in- 
tersecting this  plane  in  M.  Through  M 
draw  MG  and  MH,  at  right  angles  to  OA 
and  OB  respectively.  Pass  a  plane  through 
the  lines  P2lf  and  MG. 

Since  PM  is  perpendicular  to  OAB,  the 
plane  PMG  is  perpendicular  to  OA  B  (Euc. 
XI.  18).  Hence,  since  AGM  is  a  right 
angle,  AGP  is  also  a  right  angle.  There- 
fore  angle  PGM  Bangle  A.  Similarly  it 
can  be  shown  that  angle  PHM  —  angle  B. 


Fia...61 


FM 


PG      OPsmAOC      OP  sin  6 


.      (1) 
OP 


APPENDIX.  97 

=          CO 


.-.  by  (1)  ,  (2)  ,  sin  A  sin  5  =  sin  JS  sin  a. 

sin_B 


sin  a      sin  b 

In  a  similar  way  it  can  be  shown  that  —  —  =  sm     .     Hence 

sin  a      sin  c 

sin  A  =  sin#  _  sinC, 
sin  a      sin  &      sin  c 

Ex.  1.   Show  geometrically  : 


(a)  that  = 


sin  a      sine  sin  6       sine 

Ex.  2.    Make  the  derivation  when  M  is  not  in  the  sector  A  OB. 

4.   Analytical  derivation  of  the  law  of  sines  from  the  law  of  cosines. 

cog  A  =  cos  a-  cos  b  cos  c.  Art 

sin  &  sin  c 

.-.  1  -  cos2^  =  1  -  /  cos  a  -COB  6  cos  cV» 
\        sin  b  sin  c        ) 

_  sin2  b  sin2  c  —  cos2  a  —  cos2  b  cos2  c  4-  2  cos  a  cos  &  cos  c  . 

sin2  6  sin2  c 
_  (1  —cos2  6)(1  —  cos2  c)  —  cos2  a—  cos2  b  cos2  c+2  cos  a  cos  b  cos  c  . 

sin2  b  sin2  c 
.  .  2  j  _  1  —  cos2  a  —  cos2  b  —  cos2  c  +  2  cos  a  cos  6  cos  c 

sin2  6  sin2  c 

sin2J.  _  1  —  cos2  a  —  cos2  &  —  cos2  c  +  2  cos  a  cos  &  cos  c  ^ 

sin2  a  sin2  a  sin2  b  sin2  c 

Similarly,  sm2  B  and  sm2  ^  can  each  be  shown  to  be  equal  to  the  second 
sin2  b  sin2  c 

member  of  (1).     Hence, 

sin  A     sin  B     sin  C  2n  , 


sin  a      sin  b       sin  c      sin  a  sin  &  sin  c  ' 

in  which  2  w  denotes  the  positive  square  root  of  the  numerator  of  the  second 
member  of  (1), 

Ex.  1.    Show  the  truth  of  the  statement  made  above. 

Ex.  2.    Show  that  the  numerator  in  the  second  member  of  (1)  is  equal  to 
4  sin  s  sin  (s  —  a)  sin  (s  —  6)  sin  (s  —  c). 

A  A 

SUGGESTION,     sin  A  —  2  sin  —  cos  —  ,  and  Art.  37,  (4). 


98  SPHERICAL    TRIGONOMETRY. 

5.   Formulas  for  right-angled  triangles  derived  from  the  general  formulas. 

In  the  triangle  ABC  let  angle  (7  =  90°.  Then  sin  (7=1,  and  relations 
(1),  p.  45,  become  (2)  and  (2'),  p.  30.  Also,  cos  C  =  0,  and  the  third 
formula  in  Art.  30,  B  becomes  (1),  p.  30.  The  three  formulas  in  Art.  36,  C 
reduce  to  (5),  (5'),  and  (0),  p.  30,  respectively.  Formulas  (3),  (3'),  (4) 
and  (4'),  p.  30,  can  be  derived  from  the  others  on  that  page.  For 

cos  ^L  =  sin  .B  cos  a  [by  (5')]=—-^^  [by  (2'),  (1)1  =  —  : 

sin  c    cos  6  tan  c 

similarly, 


tan  c 
Also, 


[by  (5')]=——  [by  (2), 


, 
cosJ.     sin  B  cos  a  sin  6  cos  a  sin  b 

similarly,  tan_B  =  ^L&. 

sin  a 

Other  relations  in  triangles  (see  Art.  40)  can  also  be  used  in  the  derivation 
of  the  formulas  for  right-angled  triangles. 

EXERCISES. 

1.  Deduce  the  law  of  cosines  :  (1)  directly,  by  geometry  ;  (2)  by  means 
of  the  relations  in  a  right-angled  triangle. 

2.  Deduce  the  law  of  sines  :   (1)  analytically,  from  the  law  of  cosines 
(2)  directly,  by  geometry  ;  (3)  by  means  of  the  relations  in  a  right-angled 
triangle. 

3.  Deduce  the  ten  relations  between  the  sides  and  angles  of  a  right-angled 
spherical  triangle  :  (1)  by  means  of  the  relations  between  the  sides  and  angles 
of  the  general  spherical  triangle  ;  (2)  directly,  by  geometry. 

NOTE   B. 

[Supplementary  to  Art.  58.] 

DERIVATION    OF    FORMULAS    FOR    THE    SPHERICAL    EXCESS 
OF    A    TRIANGLE. 

J.   Cagnoli's  Formula.     (In  terms  of  the  sides.} 
BW$E  =  *m%(A  +  B  +  (7-180°)  =  -  cos$(A+  B  +  (7) 
'==  sin  l(A  +  B)  sin  \C  -  cos  \(A  +  B)  cos  \C 

=  sintCcos?C[cos  A(«  -  &)-  cos  i(«  +  &)]     [Art.  39,  ^1),  (3)] 


cos  \  c          sin  a  sin  b 

n ^ 

2  cos  £  a  cos  £  b  cos  £  c 


2n  [Note  A,  Art.  4,  Eq.  (2)] 


APPENDIX.'  '  99 

II.  Lhuillier's  Formula.     (In  terms  of  the  sides.) 


sin  KA  +  B  +  0-180°) 
0 


gin  ^  +  J)-  Bin  1080°  -C)   .  [Plane  m  94] 

cos  K^  +  -B)  +  cosK180°  -  C) 


sin  %(A  +  B}  -  cos 


sin  KS  —  6)  sin  ^(s  —  a)    /      sin  s  sin  (a  —  c) 
—  ' 


cos  £  s  cos  \  (s  —  c)        *  sin  (s  —  a)  sin  (s  —  &) 

[Art.  37,  (6);  P/a»e  TVty.,  p.  94] 

=  Vtan  £  s  tan  ^(s  —  a)  tan  ^(s  —  &)  tan  ^(s  —  c). 

III.  Formula  in  terms  of  two  sides  and  their  included  angle. 

cos  \E  =  cos  %(A  +  B  +  C  -  180°)  =  sin  %(A  +  B  +  <7) 
=  cos  ^(-4  +  5)  sin  $C  +  sin  £(.4  -f-  5)  cos  ^ (7 

=  [cos  Ka  +  6)sin2^O+  cos^(a  -  6)  cos2  JC.]  sec  ^c 

[Art.  39,  (1),  (3)] 

=  (cos  £  ci.cos  ^  &  +  sin  $  a  sin  $  6  cos (7)  sec  £  c.  (2) 

Hence,  from  (1)  and  (2),  on  division  and  reduction, 

tanAJ?=     tan  |o  tan  ^6  sinC    , 
1  +  tan  \  a  tan  \  b  cos  C 

On  taking  the  reciprocals  and  reducing,  this  takes  the  form 
.  .  „     cot  i  a  cot  \  b  +  cos  (7 

CO  I  *  Hi  = * — • 

sinC 


• 


ESTIONS   AND    EXERCISES    FOR    PRACTICE 
AND   REVIEW. 


CHAPTER  I. 

1.  On  a  sphere  let  N  be  the  pole  of  a  great  circle  ABC,  and  P  be  any 
point  on  the  surface  between  N  and  ABC;  also  let  DPNG  be  a  semicircle 
drawn  through  P  at  right  angles  to  AB C,  and  let  it  intersect  ABC  in  D 
and  Cr :  prove  (a)  that  PD  is  the  shortest  great-circle  arc  that  can  be  drawn 
from  P  to  ABC ;  (5)  that  PNG  is  the  longest  great-circle  arc  that  can  be 
drawn  from  P  to  ABC. 

2.  Show  that  the  greater  the  distance  of  the  plane  of  a  small  circle  from 
the  centre  of  the  sphere,  the  less  is  the  circle. 

3.  The  radius  of  a  sphere  is  10  inches,  and  the  radius  of  a  small  circle 
upon  it  is  6  inches.     Find :   (a)  the  distance  between  the  centre  of  the  sphere 
and  the  centre  of  the  small  circle  ;  (&)  the  angular  radius  of  the  small  circle  ; 
(c)  the  polar  distance  (or  arcual  radius)  of  the  small  circle ;   (cf)  the  dis- 
tance ,  on  the  sphere  from  the  small  circle  to  the  great  circle  having  the 
same  axis. 

4.  Prove  that  if  a  spherical  triangle  has  two  right  angles,  the  sides  oppo- 
site them  are  quadrants,  and  the  third  angle  has  the  same  measure  as  its 
opposite  side. 

5.  Prove  that  in  any  spherical  right  triangle  an  angle  and  its  opposite 
side  are  always  in  the  same  quadrant. 

6.  Prove  that  any  side  of  a  spherical  triangle  is  greater  than  the  difference 
between  the  other  two  sides. 

7.  Prove  that  each  angle  of  a  spherical  triangle  is  greater  than  the  differ- 
ence between  180°  and  the  sum  of  the  other  two  angles. 

8.  Show  that  the  surface  of  a  sphere  is  eight  times  the  surface  of  a  trirec- 
tangular  triangle: 

9.  (a)  Show  that  a  trirectangular  triangle  is  its  own  polar;   (&)  show 
that  a  triquadrantal  triangle  is  its  own  polar. 

10.    Show  that  if  two  great  circles  are  equally  inclined  to  a  third,  their 
poles  are  equidistant  from  the  pole  of  the  third. 

101 


102  SPHERICAL    TRIGONOMETRY. 

11.  Show  that  the  arc  through  the  poles  of  two  great  c  .cles  cuts  both 
circles  at  right  angles. 

12.  A  ship  sails  along  the  parallel  of  45°  N.  a  distance  of  600  nautical 
miles.     Find  the  difference  of  longitude  that  she  has  made. 

13.  Two  places  in  latitude  60°  N.  are  150  statute  miles  apart.     Find  their 
difference  of  longitude.     [Take  the  radius  of  the  earth  as  3960  miles.] 

14.  Compare  the  lengths  of  the  parallels  of  30° N.,  45° N.,  and  60° N., 
with  the  length  of  the  equator. 

15.  Prove  that  if  the  first  of  two  spherical  triangles  is  the  polar  triangle 
of  the  second,  then  the  second  is  the  polar  triangle  of  the  first. 

16.  Show  that  in  two  polar  triangles  each  angle  of  the  one  is  the  supple- 
ment of  the  side  opposite  to  it  in  the  other. 

17.  Show  that  the  sum  of  the  angles  of  a  spherical  triangle  is  greater  than 
two,  and  less  than  six,  right  angles. 

18.  Discuss  the  following  cases,  in  which  A,  a,  and  b  are  given  in  a  spheri- 
cal triangle  ABC: 

I.  ^1  =  90°:  (1)  &  =  90°;  (2)  &<90°(a<6,  a  =  b,  a>b  and  <ir-b, 
a  =  TT  -  b,  a>7r-&);  (3)  b  >90°(a<7r  -  6,  a  =  TT  -  b,  a>tr-b  and 
<6,  a  =  b,  «>&). 

II.  ^1<900:  (1)  b  =  W°(a<A,  a  =  A,  a>  A  and  <  6,  a  =  b  =  90°, 
a  >  6) ;  (2)  b  <  90°(a  <p,  a  =j>,  a  >p  and  <b,  a  =  b,  a>b  and  <TT  -  6, 
a  =  TT  —  b,  a>?r  —  6);  (3)  £>>90°(a<p,  a  =  p,  a>p  and  <TT  — 6, 
a  —  TT  —  b,  a  >  TT  —  b  and  <  b,  a  =  b,  a  >  6) .  [For  definition  of  p,  see  p.  26.] 
III.  ^4>90°:  (1)  &  =  90°(a  =  &,  a  between  &  and  TT  -  ft,  a  between 
TT  —  b  and  ^?)  ;  (2)  &<90°(a>p,  a=.p,  a<j»  and  >  &,  a  =  6,  a  between  6 
and  TT  —  6) ;  (3)  6>90°(«<&,  a>6  and  <j9,  a<p  and  >?r  — 6, 
a  between  6  and  TT  —  6,  a  =  6). 

CHAPTER  II. 

1.  Define  spherical  angle,  spherical  triangle,  Napier's  circular  parts, 
polar  triangle,  quadrantal  triangle,  oblique  spherical  triangle,  pole  of  an 
arc,  spherical  excess,  spherical  polygon. 

2.  In  a  right-angled  spherical  triangle  show  that :  (a)  It  is  impossible 
for  only  one  of  the  three  sides  to  be  greater  than  90°  ;    (6)  The  hypote- 
nuse  is  less  than   90°  only  when  both  the  other  sides   are   in  the   same 
quadrant ;  (c)  If  another  part  besides  the  right  angle  be  right,  the  triangle 
is  biquadrantal. 

3.  Prove,  by  geometry  and  by  trigonometry,  that  in  a  right  spherical 
triangle  an  angle  and  its  opposite  side  are  always  in  the  same  quadrant,  that 
is,  either  both  are  less  or  both  are  greater  than  90°. 


QUESTIONS  AND  EXERCISES.  103 

4.  Trove  tbiat  in  a  right  spherical  triangle  ABC,  (C  -  90°)  :  (a)  sin  A  = 
cos  B  +  cos  6;  (6)  cos  c  =  cot  A  cot  B ;  (c)  cos  c  =  cos  a  cos  b. 

5.  (a)  Mention  in  order  Napier's  circular  parts,  and  state  the  two  prin- 
cipal rulesr  for  their  use.    (&)  State  Napier's  Kules  and  write  the  ten  for- 
mulas for  the  right  spherical  triangle  by  means  of  them,     (c)  Prove  three 
of  these  formulas. 

6.  What  formulas  should  be  used  to  find  B,  a,  and  &  of  a  right  spherical 
triangle  ABC  (C  =  90°)  when  A  and  c  are  given  ?     What  formula  includes 
all  the  required  parts  ? 

7.  Show  how  to  obtain  the  formulas  for  finding  a,  B,  and  C  of  a  qua- 
drantal  triangle,  when  A  and  6  are  given  and  c  =  90°. 

8.  Given  one  side  and  the  hypotenuse  of  a  right  spherical  triangle,  write 
all  the  formulas  for  the  solution  and  check,  and  state  how  the  species  of  each 
part  will  be  determined. 

9.  How  many  solutions  are  there  for  a  right  spherical  triangle  ABC, 
given  side  b  and  angle  B  ?    Discuss  fully. 

10.  Given  A  and  b  of  a  right  spherical  triangle  ABC  (<7=  90°)  :  write 
and  derive  formulas  for  computing  each  of  the  parts  B,  a,  and  c  in  terms  of 
A  and  6  only  ;  also  the  check  formula. 

11.  Show  how  to  solve  a  right  spherical  triangle,  having  given  (a)  the 
sides  about  the  right  angle  ;  (&)  the  two  oblique  angles. 

12.  (a)  Show  how  the  solution  of  a  quadrantal  triangle  may  be  reduced 
to  that  of  a  right  triangle.     (&)  Write  the  relations  between  the  sides  and 
angles  of  a  quadrantal  triangle  ABC,  in  which  c  =  90°. 

13.  In  a  spherical  triangle  ABC,  A  =  B :  write  the  relations  between  the 
sides  and  angles  of  ABC. 

14.  If  A  be  one  of  the  base  angles  of  an  isosceles  spherical  triangle  whose 
vertical  angle  is  90°  and  a  the  opposite  side,  prove  that  cos  a  =  cot  A  ;  and 
determine  the  limits  within  which  it  is  necessary  that  A  must  lie. 

15.  Show  how  oblique  spherical  triangles  can  be  solved  by  means  of  right 
spherical  triangles.     (Six  cases.) 

16.  In  a  right  spherical  triangle  ABC  (C  =  90°)  prove  that:   («)  sin2  B 
—  cos2  A  —  sin2  b  sin2  A;    (&)  sin  A  sin  2  b  —  sin  c  sin  2  B ;   (c)  sin  2  A  sin  c  = 
sin  2  a  sin  B ;      (d)    sin  2  a  sin  2  b  =  4  cos  A  cos  B  sin2  c ;      (e)   cos2  A  sin2  c  = 
sin'2  c  -  sin2  a  ;  (/)  sin2  A  cos2  c  =  sin2 .4  -  sin2  a. 

17.  (a)  In  ABC,  if  C  =  90°,  and  a  =  b  =  c,  prove  that  sec  A  =  I  +  sec  «, 

(6)  In  ABC  (C=  90°)  show  that  if  b  =  c  =  -,  then  cos  a  =  cos  A. 

2 

18.  In  a  right  spherical  triangle  whose  oblique  angles  are  72°  34'  and 
59°  42',  find  the  length  of  the  perpendicular  from  the  right  angle  upon  the 
base,  and  the  angles  which  it  forms  with  the  sides. 


104  SPHERICAL    TRIGONOMETRY. 

19.  Two  planes  intersecting  at  right  angles  are  intersected  by  a  third 
plane  making  with  them  angles  of  60°  and  75°  respectively.     Find  the  angles 
which  the  three  lines  of  intersection  make  with  each  other. 

20.  Two  planes  intersect  at  right  angles ;  from  any  point  of  their  line  of 
intersection  one  line  is  drawn  in  each  plane  making  the  respective  angles  60° 
and  73°  with  the  line  of  intersection.     Find  the  angle  between  the  two  lines 
thus  drawn. 

21.  A  triangle  whose  sides  are  40°,  90°,  and  125°  respectively,  is  drawn  on 
the  surface  of  a  sphere  whose  radius  is  8  feet.    Find  in  feet  the  length  of 
each  side  of  this  triangle,  and  also  the  angles  of  the  polar  triangle.     Write 
the  formula  for  finding  either  angle  in  terms  of  functions  of  the  sides. 

22.  Solve  the  following  spherical   triangles  given:    (1)   Right  triangle, 
hypotenuse  =  140°,  one  side  =  20°.     (2)  Sides  90°,  50°,  50°.     (3)  Sides  100°, 
50°,  60°.     (4)  Sides  each  30°  in  length.     (5)  A  =  100°,  C  =  90°,  a  =  112°. 
(6)   .4  =  80°,    a  =  90°,    ft  =  37°.     (7)   a  =  b  =  119°,    (7=85°.     (8)   Triangle 
PQR,  R  =  90°,  P=  63°  42',  Q  =  123°  18'.     (9)  Right  triangle,  one  angle  = 
110°  30'  20",  hypotenuse  =  75°  45'.     (10)  A  =  90°,  ft  =  21°  30',  c  =  122°  18'. 
.(11)  5  =  90°,  C=  79° 40',  b  =137°  52'.     (12)  .4=90°,  «  =  108°23,  c=37°42'. 

(13)  #=90°,  ,4=43°  10',  a  =  78°35'.  (14)  .5=90°,  C  =33°  57',  .4=43°  18'. 
(15)  ,4  =  87°  40'  20",  b  =  33°  42'  40",  5  =  90°.  (16)  .4  =  33°  42'  40",  6  = 
87°  40'  20",  .8  =  90°. 

CHAPTER   III. 

1.  In  a  spherical  triangle  ABC  prove  that :  (a)  sin  a  :  sin  yli=sin  b  :  sin  B 
=  sin  c  :  sin  C ;       (ft)    cos  a  =  cos  ft  cos  c  +  sin  ft  sin  c  cos  A  ;      (c)    cos  A  = 
—  cos  B  cos  C+  sin  B  sin  C  cos  a  ;      (d)  cos  %A=  Vsin  s  sin(s— a)  -f-sin  ft  sin  c, 
where  s  =  £(a  +  ft  +  c) ;  (e)  tan  \  A  cot  \  B  =  sin  (s  —  ft)  cosec  (s  —  a). 

2.  Give  the  equations  (or  proportions)  known  as  Napier's  Analogies. 
Derive  them. 

3.  Derive  formulas  giving  the  values  of  sin  A,  cos  A,  tan  A,  and  cos  c,  in 
terms  of  functions  of  a,  ft,  and  c. 

4.  In  a  spherical  triangle  ABC  show  that :  (a)  If  a  =  b  =  c,  then  sec^l 
=  1  +  sec  a.     (6)  If  ft  -f  c  =  180°,  then  sin  2  B  +  sin  2  C  =  0.     (c)  If  C=  90°, 

then  tan  i(c  +  a)  tan  J(c  -  a)  =  tan2 1. 

2 

5.  In  an  equilateral  spherical  triangle  show  that :  (a)  2  sin  —  cos  |  =  1, 

Zt  -j 

and  hence,  that  such  a  triangle  can  never  have  its  angle  less  than  60°,  nor  its 
side  greater  than  120°  ;  (ft)  2  cos  .4  =  1  -  tan2  -• 

6.  Show  that :    (a)  If  the  three  angles  of  spherical  triangle  ABC  are 

together  equal  to  four  right  angles,  then  cos2  ^  =  cot  A  cot  B.     (6)  If  x  is 

2 


QUESTIONS  AND  EXERCISES.  105 

the  side  of  a  spherical  triangle  formed  by  joining  the  middle  points  of  the 

equilateral  triangle  of  side  a,  then  2  sin  -  =  tan  -• 

2  2 

7.  (a)   In  a  spherical  triangle  ABC  show  that,   if  b  -f  c  =  90°,   then 

cos  a  =  sin  2  c  cos2  — .     (6)  If  a  be  the  side  of  an  equilateral  triangle  and  a' 

2 

that  of  its  polar  triangle,  prove  cos  a  cos  a'  =  ±. 

8.  (a)  If,  in  a  triangle  ABC,  I  be  the  length  of  the  arc  joining  the  middle 
point  of  the  side  c  to  the  opposite  vertex  C,  shov^that  cos  I  =  (cos  a  +  cos  6) 
-r-  2  cos-.     (6)  In  a  right  spherical  triangle  ABC  (C  =  90°),  if  a,  ft  be  the 
arcs  drawn  from  C  respectively  perpendicular  to  and  bisecting  the  hypote- 
nuse c,  show  that  sin2  -  (1  +  sin2  a)  =  sin2  ft. 

9.  (a)  Prove  that  the  half  sum  of  two  sides  of  any  spherical  triangle  is  in 
the  same  quadrant  as  the  half  sum  of  the  opposite  angles.     ^6)  Two  sides  of 
a  spherical  triangle  are  given :  prove  that  the  angle  opposite  the  smaller  of 
them  will  be  greatest  when  that  opposite  the  larger  is  a  right  angle. 

10.  ABC  is  a  spherical  triangle  of  which  each  side  is  a  quadrant,  and  Pis 
a  point  within  it.    Prove  that  cos2  PA  +  cos2  PB  +  cos2  PC  =  1. 

11.  In  a  spherical  triangle,  if  A  =  36°,  B  =  60°,  and  C  =  90°,  show  that 
a  +  6  +  c  =  90°. 

CHAPTER  IV. 

1.  (a)  Name  the  six  cases  for  solution  of  spherical  triangles.     (6)  Dis- 
cuss each  case  in  detail,  writing  the  formulas  used  in  the  solution,  and  deriv- 
ing these  formulas,     (c)  Solve  an  example  under  each  case.     Test  the  result 
by  (1)  solving  by  right  triangles,  (2)  solving  without  logarithms,  (3)  using  a 
check  formula. 

2.  How  many  solutions  are  possible  for  the  oblique  spherical  triangle 
ABC,  given  A,  B,  and  a  ?    Discuss  in  full  the  question  of  one  solution,  two 
solutions,  or  no  solution.     Plan  the  solution. 

3.  In  a  spherical  triangle  ABC,  two  sides  a  and  b  and  the  included  angle 
C  are  given.     Write  all  the  formulas  used  in  the  solution  and  check  ;  describe 
fully  the  process  of  solution.     Derive  the  formulas  used. 

4.  Write  and  deduce  the  formulas  for  rinding  A,  B,  and  C  of  any  spheri- 
cal triangle  when  a,  b,  and  c  are  given. 

5.  Given  A,  _B,  and  C.    Show  how  to  find  the  remaining  parts,  writing 
the  formulas  to  be  used. 

6.  In  an  equilateral  spherical  triangle  the  side  a  is  given.     Find  the 
angle  A. 


106  SPHERICAL   TRIGONOMETRY. 

7.  Solve  the  spherical  triangle  whose  sides  are  70°,  60°,  and  50°.    Solve 
the  plane  triangle  obtained  by  connecting  by  straight  lines  the  vertices  of 
this  spherical  triangle,  the  sphere  on  which  it  is  drawn  being  2  feet  in 
diameter. 

8.  In  a  triangle  ABC  on  the  earth's  surface  (supposed  spherical)  a  =  483 
miles,  b  =  321  miles,  C  =  38°  21'.    Find  the  length  of  the  side  c.     [Earth's 
radius  =  3960  miles.] 

9.  Two  planes  intersect  at  an  angle  of  75°.    From  any  point  of  their  line 
of  intersection  one  line  is  drawn  in  each  plane,  making  the  respective  angles 
55°  and  80°  with  the  line  of  intersection.    Find  the  angle  between  the  lines 
thus  drawn. 

10.  Two  planes  intersecting  at  an  angle  of  65°  are  intersected  by  a  third 
plane,  making  with  them  the  respective  angles  55°  and  82°.     Find  the  angles 
which  the  three  lines  of  intersection  make  with  one  another. 

11.  A  solid  angle  is  contained  by  three  plane  angles  62°,  83°,  38°.     Find 
the  angle  between  the  planes  of  the  angles  62°  and  38°. 

12.  Two  of  the  three  angles  which  contain  a  solid  angle  are  42°  and  65°  30', 
and  their  planes  are  inclined  at  an  angle  of  50°.     Find  the  angle  of  the  third 
plane  face  and  the  angles  at  which  this  third  plane  is  inclined  to  the  other 
two  planes. 

13.  A  pyramid  has  each  of  its  slant  sides  and  base  an  equilateral  triangle. 
Find  the  angle  between  any  two  faces. 

14.  A  pyramid  each  of  whose  slant  faces  is  an  equilateral  triangle  has  a 
square  base.     Find  the  angle  between  any  two  slant  faces,  also  the  angle 
between  any  slant  face  and  the  base. 

15.  In  the  following  cases  ABC  is  a  three-sided  spherical  figure  each  of 
whose  sides  is  an  arc  of  a  great  circle.     Select  those  which  are  spherical  tri- 
angles, and  give  reasons  for  so  doing.     Explain  why  the  other  figures  can- 
not be  triangles.    Solve  the  triangles  and  check  the  results.     (Solve  some 
without  using  logarithms.) 

(1)  a  =  76°,  b  =  54°,  c  =  36°.  (2)  A  =  54°  35'  20",  b  =  104°  25'  45", 

c  =  92°10'.  (3)  A  =  107-°  47' 7",   B  =  38°  58'  27",    c=  51°  41'  14". 

(4)  A  =  60°,   B  =  80°,    C  =  100°.  (5)  A  =  120°,  B  =  130°,   C  =  80°. 

(6)  .4=54°  35',  6  =  104°  24',  c=95°  10'.  (7)  .4=61°  37'  53",  B=  139°  54'  34", 
6  =  150°  17' 26".  (8)  a  =  72°18',  b  =146°  35',  c=98°ll'.  (9)  .4  =  125°  15', 
C= 85°  12',  6  =  100°.  (10)  .4=50°,  5=114°  5'  8",  6  =  50°.  (11)  ^1=83°  40', 
b  =  73°  45',  a  =  30°  24'.  (12)  A  =  83°  40',  b  =  30°  24',  a  =  73°  45'. 

(13)  A  =  97°  20',  a  =  94°  37',  b  =  36°  17'.  (14)  a  =  127°  40',  b  =  143°  50', 
c  =  139°  39*.  (15)  A  =  40°,  B  =  30°,  C  =  20°.  (16)  A  =  40°  35', 

B  =  36°  42',  c  =  47°18'. 


QUESTIONS  AND  EXERCISES.  107 


CHAPTER  V. 

[In  the  following  exercises, 


w  =  Vsin  s  sin  (s  —  a)  sin  (s  —  &)  sin  (s  —  c), 
and  N  =  V-  cos  #cos  (S  -  -4)  cos  (£-  B}  cos  (#  -  C)  ; 

also,  7%  ra,  7*5  rc,  denote  the  radii  of  the  circles  inscribed  in  the  spherical  tri- 
angle ABC  and  its  three  colunar  triangles,  and  jf?,  _Z?a,  Rb,  Rc  denote  the 
radii  of  the  circumscribing  circles  of  these  triangles.] 

1.  Given  a  spherical  triangle.  AB  C,  find  (1)  the  radius  of  the  inscribed 
circle  ;  (2)  the  radius  of  the  circumscribing  circle  ;  (3)  the  radii  of  the 
inscribed  circles  of  the  colunar  triangles ;  (4)  the  radii  of  the  circumscribing 
circles  of  the  colunar  triangles. 

Show  that : 

sins 

N 


3. 


cos  (S  -  A)  cos  (S  -  B)  cos  (#  -  C) 
4.    (a)  Cot  R  cot  Ra  cot  Rb  cot  Rc  =  N2  ; 
(6)  Tan  R  cot  Ra  cot  Rb  cot  Jf?c  =  cos2  S. 
—  cos  $  sin  s 


5.    Tan  .R  =  4  tan  r 


sin  a  sin  &  sin  c  sin  A  sin  B  sin  (7 

6.  Tan  ra  tan  r&  tan  rc  =  tan  r  sin2  s. 

7.  Tan  R  -f  cot  r  =  tan  JKa  -f  cot  ra  =  tan  J?5  +  cot  rb 

=  tan  Rc  +  cot  rc  =  |  (cot  r  -f  cot  ra  +  cot  r&  +  cot  rc). 

8.  Tan^tanr  =  -COS^sina^-cos-ysln6^etc.    Write  the  other  for- 


mulaofthisset.  sins  sin  5 

9.    Tan2  j£  +  tan2  Ra  +  tan2  Rb  +  tan2  J?c  =  cot2  r  +  cot2  ra  +  cot2  rb  +  cot2  rc. 

10.  Tan  r  tan  ra  tan  r6  tan  rc  =  w2  ;   cot  r  tan  ra  tan  rt  tan  rc  =  sin2  s. 

11.  In  any  equilateral  triangle,  tan  R  =  2  tan  r. 

12.  TanJgB=     tan^  =  cos(^-^)^  sin^a 

cos  jS  N  sin  ^.  sin  \  b  sin  ^  c 

_  &)  +  gin(s  ^ 


7i  2  71 

Write  the  corresponding  formulas  for  Rb  and  7?c. 

13.  Cot  ra  +  cot  r&  +  cot  re  —  cot  r  =  2  tan  .R. 

14.  Find  the  radii  of  the  circles  connected  with  some  of  the  triangles  -in 
Ex.  15  of  the  preceding  set. 


108  SPHERICAL   TRIGONOMETRY. 


CHAPTER   VI. 

1.  Define  the  following  terms  :  zone  of  a  sphere,  lune,  spherical  degree, 
spherical  excess  of  a  triangle,  spherical  excess  of  a  (non-re-entrant)  polygon, 
spherical  excess  of  any  figure  on  a  sphere,  spherical  measure  and  spherical 
degree  measure  of  a  solid  angle,  spherical  pyramid,  spherical  sector,  spherical 
segment. 

2.  Derive  the  area  of  the  surface  of  a  sphere. 

3.  Derive  the  area  of  a  spherical  triangle. 

4.  Discuss  fully  the  measurement  of  solid  angles. 

5.  Show  how  to  find  the  spherical  excess  of  a  figure  on  a  sphere  when 
the  area  of  the  figure  is  given  (in  square  units). 

6.  State  and  deduce  Roy^s  Rule  for  computing  the  spherical  excess  of  a 
triangle  of  known  area  on  the  earth's  surface. 

7.  Derive  the  volumes  of  a  sphere,  a  spherical  pyramid,  a  spherical 
sector,  and  a  spherical  segment. 

8.  The  area  of  an  equilateral  triangle  is  one-fourth  the  area  of  the 
sphere :   find  its  sides  and  angles. 

9.  If  the  three  sides  of  a  spherical  triangle  measured  on  the  earth's 
surface  be  12,  16,  and  18  miles,  find  the  spherical  excess. 

10.  If  a  =  b  and  C  =  -    show  that  tan#°  =  sm2  a  .    (jn  ABC.) 

2'  2  cos  a 

11.  If  a  =  b  =  60°  and  c  =  90°,  show  that  E  =  cos-1  $.    (In  ABO.) 

12.  If  C  =  90°  in  ABC,  then  E  =  2  tan-1  (tan  £  a  tan  \  6). 

13.  In  a  triangle  on  the  earth's  surface  (assumed  spherical),  two  sides 
are  483  and  321  miles,  and  the  angle  between  them  is  38°  21'.     Find  the  area 
of  the  triangle  in  square  miles.     [Radius  of  earth  =  3960  miles.] 

14.  The  sides  of  a  triangle  on  the  earth's  surface  (supposed  spherical) 
are  321,  287,  and  412  miles  ;  find  the  area. 

15.  Prove  that  in  a  right  triangle  ABC  (C  =  90°), 

cosiff  =  cos*qcosa6,  and   8iniJr  =  ggJg8in'*6. 
cos  \c  cos  I  c 

16.  The  spherical  excess  of  a  triangle  on  the  earth's  surface  is  2". 6. 
Find  its  area,  the  radius  of  the  earth  being  taken  as  3960  miles. 

17.  Find  the  fraction  of  the  earth's  surface  (supposed  spherical)  con- 
tained by  great-circle  arcs  joining  London,  New  York,  and  Paris.     Find  the 
spherical  degree  measure,  and  the  spherical  measure  of  the  angle  subtended 
at  the  centre  of  the  earth  by  this  part  of  the  earth's  surface. 


QUESTIONS  AND  EXERCISES.  109 

18.  Find  the  spherical  excess  of  some  of  the  triangles  in  Ex.  15,  p.  104. 
Also  find  their  areas  in  square  inches  on  spheres  of  radii,  say,  4  inches, 
10  inches,  12  inches,  20  inches,  a  inches. 

19.  Find  the  spherical  measures  and  the  spherical  degree  measures  of  the 
solid  angles  corresponding  to  the  triangles  taken  in  Ex.  18. 


CHAPTER  VII. 

1.  Given  the  latitude  and  longitude  of  each  of  two  places :  show  how  to 
find  the  shortest  distance  between  these  places,  and  the  direction  of  one  place 
from  the  other. 

2.  Given  the  latitudes  and  longitudes  of  three  places  on  the  earth's  sur- 
face, and  also  the  radius  of  the  earth :  show  how  to  find  the  area  of  the 
spherical  triangle  formed  by  arcs  of  great  circles  passing  through  them. 

3.  Given  the  sun's  altitude  and  declination  and  the  latitude  of  a  place : 
show  clearly  how  the  time  of  day  may  be  determined. 

4.  If  d  represents  the  sun's  declination,  what  formulas  will  be  required  in 
order  to  determine  the  time  of  sunrise  for  a  place  whose  latitude  is  I  ? 

5.  Show  what  formulas  must  be  used  to  find  the  length  of  a  degree  of 
longitude  on  the  earth's  surface  for  a  place  whose  latitude  is  Z,  r  representing 
the  radius  of  the  earth. 

6.  The  shortest  distance  d  between  two  places  and  their  latitudes  I  and  I' 
are  known ;  find  their  difference  of  longitude. 

7.  Given  the  obliquity  of  the  ecliptic  w,  and  the  sun's  longitude  X,  show 
that  if  a  and  d  denote  his  right  ascension  and  declination  respectively,  then 
tan  a  =  cos  a>  tan  X,  and  sin  S  =  sin  w  sin  X. 

8.  The  faces  of  a  regular  dodecaedron  are  regular  pentagons,  three  faces 
meeting  at  each  vertex.     Find  the  diedral  angle  at  the  edge  of  the  solid. 

9.  The  ridges  of  two  gable  roofs  meet  at  right  angles ;  each  roof  is 
inclined  to  the  horizontal  at  an  angle  of  65°.     Find  the  diedral  angle  between 
the  planes  of  the  two  roofs,  and  the  angle  their  line  of  intersection  makes 
with  the  ridge  of  either  roof. 

10.  What  is  the  direction  of  a  wall  in  latitude  52°  30'  N.  which  casts  no 
shadow  at  6  A.M.  on  the  longest  day  of  the  year  ? 

11.  Two  ports  are  in  the  same  parallel  of  latitude,  their  common  latitude 
being  J,  arid  their  difference  of  longitude  2  X.     Show  that  the  saving  of  dis- 
tance in  sailing  from  one  to  the  other  on  the  great  circle  instead  of  sailing 
due  east  or  west,  is 

2  r  {X  cos  I  —  sin~1(sin  X  cos  Z)}, 

\  being  expressed  in  radian  measure,  and  r  being  the  radius  of  the  earth. 


110  SPHERICAL    TRIGONOMETRY. 

12.  If  a  ship  sails  from  New  York  (40°  28'  N. ,  74°  8'  W.)  starting  due  east, 
and  continues  her  course  on  an  arc  of  a  great  circle,  what  will  be  her  lati- 
tude when  she  reaches  the  meridian  of  Greenwich,  and  in  what  direction  will 
she  then  be  sailing  ? 

13.  Find  the  distance  between  New  York  (40°  28'  N.,  74° 8'  W.)  and  Cape 
Clear  (51°  26'  N.,  9°  29'  W.),  and  the  bearing  of  each  from  the  other.    [Radius 
of  earth  =  3960  miles.] 

14.  From  Victoria,  B.C.  (48°25'N.,  123° 23'  W.),  a  ship  sails  on  an  arc 
of  a  great  circle  for  1250  miles,  starting  in  the  direction  S.  47°  35'  W.     Find 
its  latitude  and  longitude,  taking  the  length  of  1°  as  69|  miles. 

15.  Two  places  are  both  in  latitude  50°  N.,  and  the  difference  of  their 
longitudes  is  60°.     Find  the  distance  between  them  (a)  along  the  parallel  of 
latitude,  (&)  along  a  straight  line,  (c)  along  a  great  circle.     [Earth's  radius 
=  3960  miles.] 

16.  What  will  be  the  first  course  and  the  shortest  (great  circle)  distance 
passed  over  in  sailing  from  a  place  in  latitude  43°  N.  to  another  place  80° 
east  of  it  and  in  the  same  latitude  ?    What  is  the  distance  between  the  two 
places  along  the  parallel  ?     What  is  the  straight-line  distance  between  them  ? 

17.  At  what  hours  will  the  sun  rise  in  London  (51°  30'  48"  N.)  and  New 
York  (40°  43'  N.)  when  its  decimation  is  respectively  23°  N.,  20°  N.,  15°  N., 

10°  N.,  5°  N.,  5°  S.,  10°  S.,  15°  S.,  20°  S.,  23°  S.  ? 

18.  When  the  sun's  declination  is  18°,  find  his  right  ascension  and 
longitude. 

19.  What  is  the  altitude  of  the  sun  above  the  horizon  when  its  angular 
distance  from  the  south  point  is  75°  and  from  the  west  point  is  60°  ? 

20.  The  right  ascension  of  Sirius  is  6h  38m  37S.6,  and  his  declination  is 
16°  31'  2"  S. ;  the  right  ascension  of  Aldebaran  is  4h  27m  258.9,  and  his  decli- 
nation is  16°  12'  27"  N.     Find  the  angular  distance  between  these  stars. 

21.  If  the  sun's  declination  be  20°  45'  N.  and  his  altitude  be  41°  10'  at 
3  P.M.,  find  the  observer's  latitude. 

22.  What  will  be  the  altitude  of  the  sun  at  3.30  P.M.  in  San  Francisco 
(37° 48'  N.),  its  declination  being  15°  S.  ? 

23.  In  Bombay  (18°  54'  N.)  the  altitude  of  the  sun  is  observed  to  be 
27°  40'.     If  the  sun's  declination  is  7°  S.  and  the  observation  is  made  in  the 
morning,  find  the  hour  of  the  day. 

24.  Find  the  latitude  and  longitude  of  a  star  whose  right  ascension  is 
4h  40™,  and  declination  57°. 

25.  Find  the  distance  in  degrees  between  the  sun  and  moon  when  their 
right  ascensions  are  respectively  15h  12',  4h  45',  and  their  declinations  are 
21°  30'  S.,  5°30'N. 


QUESTIONS  AND  EXERCISES.  Ill 

26.  Find  the  length  of  the  longest  day  in  the  year  at  the  following  places 
(the  sun's  greatest  declination  being  23°  27'  N.)  :  London  (51°  30' 48"  N.), 
New  York  (40°  43'  N.),  Montreal  (45°  30'  N.),  St.  Petersburg  (60°  N.),  Hong 
Kong  (22°17'N.). 

27.  Find  the  length  of  the  shortest  day  in  the  year  at  the  places  mentioned 
in  Ex.  26.     (The  sun's  declination  is  then  23°  27'  S.) 

28.  At  Copenhagen  (55°  40'  N.),  at  an  afternoon  observation,  the  sun's 
altitude  is  44°  20' ;  find  the  time  of  day,  the  sun's  declination  being  18°  25'  N. 

29.  At  what  time  of  day  will  the  sun  have  an  altitude  of  53°  40'  for  a 
place  in  latitude  40°  35'  N.,  his  declination  being  13° 48'  N.  ? 

30.  What  will  be  the  sun's  altitude  at  3.30  P.M.  at  a  place  in  latitude 
44° 40'  N.,  his  declination  being  18°  N.  ? 

31.  What  will  be  the  sun's  altitude  at  10  A.M.  at  a  place  in  latitude 
44°  40'  N.,  his  decimation  being  18°  S.  ? 

32.  What  is  the  sun's  declination  when  his  altitude  at  a  place  in  latitude 
37°48'N.  is  25°  at  4  P.M.  ? 

NOTE.     The  Spherical  Trigonometries  of  M'Clelland  and  Preston,  Casey, 
and  Bowser,  contain  especially  good  collections  of  exercises.    See  Art.  40, 


ANSWERS   TO   THE   EXAMPLES. 


CHAPTER   I. 

Art.  24.     I.   4.    A  =  88°  12.2',  B  =  74°  34.7',  C  =  43°  8' ;  A  =  118°  33.2', 

#  =  113°  11. 2',  C=92°45'.       II.    4.    a  =  72°  40.6',  6=67°  45.8',  c  =  51°  43.1'; 
a=71°22.7',  6  =  108°  37.3',  c  =  104°56.7'.     III.  4.  ^L=63°56',  £=126°  21.2', 
c  =  77°3';  £  =  32°  47.1',  0=  62°  30.7',  a  =  84°  29.5'.       IV.    5.    6  =  70°  5.7', 
c  =  102°51.3',     ^1  =  68°  35.8';     a  =  46°  1.5',     c  =  86°  0.7',     B  =  122°  55.8'. 
VI.    3.   .B  =  59°  40.1',     <7=114°55',     c  =  96°  31.1',     and    B  =  120°  19.9', 
C  =  27°  49.6',     c  =  30°  45.4';      J?  =  65°  1.8',      C  =  97°  16.9',      c  =  100°  26'; 
C=  110°  43.1',     b  =  33°  8.6',     c  =  60°  28.8';      C  =  165°  3.3',     6  =  125°  1.7', 
c  =  162°  55.7',  and  C  =  119°  47',  c  =  81°  7',  b  =  54°  58.3'. 

CHAPTER   II. 

Art.  27.  4.  c  =  82°  33.9',  ^  =  60°  51.2',  B  =  76°  56.1'.  5.  a=33°0.25', 
b  =  36°  29.4',  c=47£37. 8'. 

Art.  31.  5.  (!>#*=  86°  30.9',  .4=36°  30.2',  5=87°  25.4'.  (2)  6  =  138° 24.4', 
4  =  68°  41.9',  B  =129°  43.1'.  &*)  a  =  35°  50.6',  6  ^.75°  39. 5',  B  =  Sl°  29.1'. 
(tf  a  =  42°  49.8',  b  =  27°  47.3',  c  =  49°  33'.  &£b  =  33°  37.4',  c  =  79°  2', 
_B=34°20.1';  and  6  =  146° 22.6',  c=100°58',  J?=145° 39.9'.  (J6)  a =35°  16.4', 
c  =  51°  10.8',  5  =  55°  18.6'.> 

Art.  32.  1.  (1)  fc  =  54°20',  ^=32°0.75',  J?  =  57°59.25',  (7=93°59.3'; 
(2)  6  =  66°  29',  c  =  lll°29.4',  jB=50°17',  O=128°41.2'.  2.  (1)  6  =  59°  56.2', 
A  =  130°,  B  =  52°  55.5'.  (2)  a  =  135°  33',  b  =  100°  58.6',  C  =  101°  24.7'. 

CHAPTER   III. 

Art.  37.  I.  2.  ^1=55° 58.4',  J5=74°  14.6',  (7=103°  36.6'.  3.  A =43° 58', 
B  =  58°  14.4',  G  =  108°  4.8'.  II.  3.  a  =  39°  29.6',  b  =  35°  36.2',  c  =  27°  59'. 
4.  a  =  130°  49.6',  6  =  120°  17.5',  c  =  54°  56.1'. 

CHAPTER  IV. 

Art.  42.  2.  A  =  41°  27',  B  =  66°  26.4',  G  =  106°  3,2'.  3.  A  =  144°  26.6', 
^  =  26°  9.1',  C=  36°  34. 7'.' 

Art.  43.  1.  a  =  43°  36',  b  =  41°  20.9','  c  =  33°  7.4'.  2.  a  =  111°  40.2', 

*  =  91°  17.2',  c  =  71°  7.4'. 

113 


114  ANSWERS   TO   THE  EXAMPLES. 

Art.  44.  2.  A  =  101°  21.2',  B  =  54°  57.9',  c  =  79°  9.5'.  3.  B  =  78°  20.6', 
<C  =  47°47',  a  =  82°  42'. 

Art.  45.  1.  a  =  63°  15.1',  b  =  43°  53.7',  C=  95°  1'.  2.  6  =  86°  39.5', 
c  =  68°  39.5',  A  =  59°  44'. 

Art.  46.  2.  B  =  36°  35.5',  C  =  51°  59.7',  c  =  42°  38.9'.  3.  B  =  59°  3.5', 
C  =  97°  38.8',  c  =  56°  56.9' ;  B  =  120°  56.5',  C  =  28°  5.2',  c  =  23°  27.8'. 

Art.  47.  1.6  =  154°  45.1',  c  =  34°  9.1',  C  =  70°  17.5'.  2.  ^1  =  164°  43.7', 
a  =  162°  37.5',  c  =  124°  40.6' ;  A  =  119°  18.7',  a  =  81°  18.7',  c  =  55°  19.4'. 

CHAPTEE   VI. 

Art.  53.     1.   2827.44  sq.  in.        2.    392.7  sq.  in.        3.    8.25  sq.  ft. 

Art.  55.     1.    1.396  sq.ft.         2.    64.14  sq.  ft. 

Art.  56.  24°  37' 47"  (.42986),  33°  56.6'  (.59213),  27°  10.4'  (.47426), 
12°  (.20944),  86° 20'  (1.5068),  etc. 

Art.  57.  1.  42.986  sq.  ft.,  59.213  sq.  ft.,  47.426  sq.  ft.  2.  130.9  sq.  in., 
941.75  sq.  in. 

Art.  61.  1.  Spherical  degree  measure  =  12,  spherical  measure  =  .20944, 
2.  Spherical  degree  measure  =  24.63,  spherical  measure  =  .42986. 

Art.  64.  1.  143.29  cu.  ft.,  197.38  cu.  ft.,  158.09  cu.  ft.,  1090.8  cu.  in., 
7847.9  cu.  in.,  etc.  2.  (a)  1357.17  cu.  in.  (6)  904.78  cu.  ft. 

CHAPTER   VII. 

Art.  66.  2.  8°  4.3' S. ;  course,  S.  45°  6  E.  5.  (a)  On  the  equator  in 
long.  18°  56'  E. ;  course,  S.  47°  39'  E.  (6)  Lat.  42°  21'  S.,  long.  108°  56'  E. ; 
course,  E.  (c)  On  the  equator  in  long.  161°  4'  W.  ;  course,  N.  47°  39'  E. 
6.  Distance  =(51°  19.8')=  3547.675  mi. ;  bearing  of  New  York  from  Liver- 
pool is  N.  71°  6.8'  W.,  and  bearing  of  Liverpool  from  New  York  is  N.  48°  5.8'  E.  ; 
lat.  51°  441'  N. ;  course,  N.  65°  38'  E. 

Art.  73.  1.  8.08A.M.  2.  2.33P.M.  3.  2.59  P.M.  4.  4.09P.M. 
5.  6.09  P.M.  6.  9.46  A.M. 

Art.  74.  1.  (a)  5.44;  (6)  5.06;  (c)  4.34;  (<*)  5.43;  (e)  6.21; 
(/)  6.53;  (0)  7.26.  2.  (a)  5.37;  (6)  4.40;  (c)  3.51;  (d)  5.35; 
(e)  6.30;  (/)  7.19;  (0)  8.09.  .  3.  (a)  5.29;  (6)  4.08;  (c)  2.51; 
(d)  5.25;  (e)  6.42;  (/)  7.51.  (j) 


14  DAY  USE 

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